255 Compiled And Solved Problems In Geometry And Trigonometry

Florentin Smarandache
Compiled and Solved Problems
in Geometry and Trigonometry

255 Compiled and Solved Problems in Geometry and Trigonometry
1
FLORENTIN SMARANDACHE
255 Compiled and Solved Problems
in Geometry and Trigonometry
(from Romanian Textbooks)
Educational Publisher
2015

2
Peer reviewers:
Prof. Rajesh Singh, School of Statistics, DAVV, Indore (M.P.), India.
Dr. Linfan Mao, Academy of Mathematics and Systems, Chinese Academy of Sciences,
Beijing 100190, P. R. China.
Mumtaz Ali, Department of Mathematics, Quaid-i-Azam University, Islamabad, 44000,
Pakistan
Prof. Stefan Vladutescu, University of Craiova, Romania.
Said Broumi, University of Hassan II Mohammedia, Hay El Baraka Ben M'sik,
Casablanca B. P. 7951, Morocco.
E-publishing, Translation & Editing:
Dana Petras, Nikos Vasiliou
AdSumus Scientific and Cultural Society, Cantemir 13, Oradea, Romania
Copyright:
Florentin Smarandache 1998-2015
Educational Publisher, Chicago, USA
ISBN: 978-1-59973-299-2

3
Table of Content
Explanatory Note..................................................................................................................................................... 4
Problems in Geometry (9th
grade)................................................................................................................... 5
Solutions ...............................................................................................................................................................11
Problems in Geometry and Trigonometry .................................................................................................38
Solutions ...............................................................................................................................................................42
Other Problems in Geometry and Trigonometry (10th
grade)..........................................................60
Solutions ...............................................................................................................................................................67
Various Problems...................................................................................................................................................96
Solutions ...............................................................................................................................................................99
Problems in Spatial Geometry...................................................................................................................... 108
Solutions ............................................................................................................................................................ 114
Lines and Planes ................................................................................................................................................. 140
Solutions ............................................................................................................................................................ 143
Projections............................................................................................................................................................. 155
Solutions ............................................................................................................................................................ 159
Review Problems................................................................................................................................................. 174
Solutions ............................................................................................................................................................ 182

4
Explanatory Note
This book is a translation from Romanian of "Probleme Compilate şi Rezolvate de
Geometrie şi Trigonometrie" (University of Kishinev Press, Kishinev, 169 p., 1998), and
includes problems of 2D and 3D Euclidean geometry plus trigonometry, compiled and
solved from the Romanian Textbooks for 9th and 10th grade students, in the period
1981-1988, when I was a professor of mathematics at the "Petrache Poenaru" National
College in Balcesti, Valcea (Romania), Lycée Sidi El Hassan Lyoussi in Sefrou (Morroco),
then at the "Nicolae Balcescu" National College in Craiova and Dragotesti General
School (Romania), but also I did intensive private tutoring for students preparing their
university entrance examination. After that, I have escaped in Turkey in September 1988
and lived in a political refugee camp in Istanbul and Ankara, and in March 1990 I
immigrated to United States. The degree of difficulties of the problems is from easy
and medium to hard. The solutions of the problems are at the end of each chapter.
One can navigate back and forth from the text of the problem to its solution using
bookmarks. The book is especially a didactical material for the mathematical students
and instructors.
The Author

5
Problems in Geometry (9th
grade)
1. The measure of a regular polygon’s interior angle is four times bigger than
the measure of its external angle. How many sides does the polygon have?
Solution to Problem 1
2. How many sides does a convex polygon have if all its external angles are
obtuse?
3. Show that in a convex quadrilateral the bisector of two consecutive angles
forms an angle whose measure is equal to half the sum of the measures of
the other two angles.
4. Show that the surface of a convex pentagon can be decomposed into two
quadrilateral surfaces.
5. What is the minimum number of quadrilateral surfaces in which a convex
polygon with 9, 10, 11 vertices can be decomposed?
6. If ̂ ≡ ′ ′ ′
̂ , then bijective function = ̂ → ′ ′ ′
̂ such
that for ∀ 2 points , ̂ , ‖ ‖ = ‖ ‖, ‖ ‖, and vice versa.

6
7. If ∆ ≡ ∆ ′ ′ ′
then bijective function = → ′ ′ ′
such that
∀ 2 points , , ‖ ‖ = ‖ ‖, ‖ ‖, and vice versa.
8. Show that if ∆ ~∆ ′ ′ ′
, then [ ]~[ ′ ′ ′].
9. Show that any two rays are congruent sets. The same property for lines.
10. Show that two disks with the same radius are congruent sets.
11. If the function : → ′
is isometric, then the inverse function −
: → ′
is as well isometric.
12. If the convex polygons = , , … , and ′
= ′
, ′
, … , ′
have | , + | ≡
| ′
, +
′
| for = , , … , − , and �+ �+
̂ ≡ ′
�+
′
�+
′
̂ , ∀ = , , … , −
, then ≡ ′
and [ ] ≡ [ ′].
13. Prove that the ratio of the perimeters of two similar polygons is equal to
their similarity ratio.
14. The parallelogram has ‖ ‖ = , ‖ ‖ = and = . Find
, .

7
15. Of triangles with ‖ ‖ = and ‖ ‖ = , and being given
numbers, find a triangle with maximum area.
16. Consider a square and points , , , , , , , that divide each side
in three congruent segments. Show that is a square and its area is
equal to �[ ].
17. The diagonals of the trapezoid || cut at .
a. Show that the triangles and have the same area;
b. The parallel through to cuts and in and . Show that
|| || = || ||.
18. being the midpoint of the non-parallel side [ ] of the trapezoid ,
show that �[ ] = �[ ].
19. There are given an angle ̂ and a point inside the angle. A line
through cuts the sides of the angle in and . Determine the line
such that the area ∆ to be minimal.
20. Construct a point inside the triangle , such that the triangles ,
, have equal areas.
21. Decompose a triangular surface in three surfaces with the same area by
parallels to one side of the triangle.

8
22. Solve the analogous problem for a trapezoid.
23. We extend the radii drawn to the peaks of an equilateral triangle inscribed
in a circle , , until the intersection with the circle passing through the
peaks of a square circumscribed to the circle , . Show that the points
thus obtained are the peaks of a triangle with the same area as the
hexagon inscribed in , .
24. Prove the leg theorem with the help of areas.
25. Consider an equilateral ∆ with ‖ ‖ = . The area of the shaded
surface determined by circles , , , , , is equal to the area of
the circle sector determined by the minor arc ̂ of the circle , .
26. Show that the area of the annulus between circles , and , is
equal to the area of a disk having as diameter the tangent segment to
circle , with endpoints on the circle , .
27. Let [ ], [ ] two ⊥ radii of a circle centered at [ ]. Take the points and
on the minor arc ̂ such that ̂ ≡ ̂ and let , be the projections of
onto . Show that the area of the surface bounded by [ ], [ [ ]]
and arc ̂ is equal to the area of the sector determined by arc ̂ of the
circle , ‖ ‖ .

9
28. Find the area of the regular octagon inscribed in a circle of radius .
29. Using areas, show that the sum of the distances of a variable point inside
the equilateral triangle to its sides is constant.
30. Consider a given triangle and a variable point | |. Prove that
between the distances = , and = , is a relation of +
= type, where and are constant.
31. Let and be the midpoints of sides [ ] and [ ] of the convex
quadrilateral and { } = and { } = . Prove that the
area of the quadrilateral is equal to the sum of the areas of triangles
and .
32. Construct a triangle having the same area as a given pentagon.
33. Construct a line that divides a convex quadrilateral surface in two parts
with equal areas.
34. In a square of side , the middle of each side is connected with the ends of
the opposite side. Find the area of the interior convex octagon formed in
this way.

10
35. The diagonal [ ] of parallelogram is divided by points , , in 3
segments. Prove that is a parallelogram and find the ratio between
�[ ] and �[ ].
36. There are given the points , , , , such that = { }. Find the
locus of point such that �[ ] = �[ ].
37. Analogous problem for || .
38. Let be a convex quadrilateral. Find the locus of point inside
such that �[ ] + �[ ] = , – a constant. For which values of the
desired geometrical locus is not the empty set?

11
Solutions
Solution to Problem 1.
−
=  =
Let =
, , ∢ ext
>
>
>
} + + > , so = is possible.
Let =
, , , ∢ ext
>
>
} + + + > , so = is impossible.
Therefore, = .

12
m(̂) =
m(̂) + m ̂
m( ̂) + m( ̂) + m( ̂) + m ̂ = °
m( ̂) + m( ̂)
= ° −
m( ̂) + m ̂
m(̂) = ° −
m ̂
−
m ̂
=
= ° − ° +
m( ̂) + m ̂
=
m( ̂) + ̂
Let ̂ ⇒ , int. ̂. Let | | ⇒ int. ̂ ⇒ | ⊂ int. ̂, | |
| = ⇒ quadrilateral. The same for .
9 vertices; 10 vertices; 11 vertices;
4 quadrilaterals. 4 quadrilaterals. 5 quadrilaterals.

13
We assume that ABC
̂ ≡ A’B’C’
̂ . We construct a function :
̂ → ′ ′ ′
̂ such that
{
B = B′
if P |BA, P B′A′
| , ′ ′
such that ‖ ‖ = ‖ ′ ′‖ where ′
= .
The so constructed function is bijective, since for different arguments there are
different corresponding values and ∀ point from ′ ′ ′ is the image of a single
point from ̂ (from the axiom of segment construction).
If , this ray,
‖BP‖ = ‖B′
P′‖
‖BQ‖ = ‖B′Q′‖
} ‖PQ‖ = ‖BQ‖ − ‖BP‖ = ‖B′Q′‖ − ‖B′
P′‖ = ‖P′Q′‖ = ‖ P , Q ‖.
If , a different ray,
‖BP‖ = ‖B′
P′‖
‖BQ‖ = ‖B′Q′‖
PBQ
̂ ≡ P′B′Q
̂ ′
} ∆PBQ = ∆P′
B′
Q′
‖PQ‖ = ‖P′
Q′‖ = ‖ P , Q ‖.
Vice versa.
Let ∶ ABC → A′
B′
C′
such that bijective and ‖PQ‖ = ‖ P , Q ‖.
Let , | and | .

14
‖PQ‖ = ‖P′
Q′‖
‖PS‖ = ‖P′
S′‖
‖QS‖ = ‖Q′S′‖
} ∆PQS ≡ ∆P′
Q′
S′
QPS
̂ ≡ Q′P′S′
̂ BPS
̂ ≡ B′P′S′
̂ (1);
‖ ‖ = ‖ ′ ′‖
‖ ‖ = ‖ ′ ′‖
‖ ‖ = ‖ ′ ′‖
} ∆PRS ≡ ∆P′R′S′′
PSB
̂ ≡ P′S′B′
̂ (2).
From (1) and (2) PBC
̂ ≡ P′B′S′
̂ (as diff. at 180°) i.e. ABC
̂ ≡ A′B′C′
̂ .
Let ∆ ≡ ∆ ′ ′ ′.
We construct a function ∶ → ′ ′ ′ such that = ′, = ′, = ′
and so | | → ′ = | ′ ′| such that || || = || ′ ′||;
| | → ′ = | ′ ′| such that || || = || ′ ′||;
| | → ′ = | ′ ′| such that || || = || ′ ′||.
The so constructed function is bijective.
Let | | and | | ′ | ′ ′| and ′ | ′ ′|.
‖AP‖ = ‖A′P′‖
‖CQ‖ = ‖C′Q′‖
‖CA‖ = ‖C′A′‖
} ‖AQ‖ = ‖A′Q′‖; A ≡ A′
∆APQ ≡ ∆A′P′Q′ ‖PQ‖ = ‖P′Q′‖.
Similar reasoning for (∀) point and .
Vice versa.
We assume that a bijective function ∶ → ′ ′ ′ with the stated
properties.
We denote A = A′′
, B = B′′
, C = C′′
‖AB‖ = ‖A′′
B′′‖, ‖BC‖ = ‖B′′
C′′‖, ‖AC‖ = ‖A′′
C′′‖∆ABC = ∆A′′
B′′
C′′
.
Because ABC = [AB] [BC] [CA] = [AB] [BC] [CA]
= [A′′B′′] [B′′
C′′][C′′
A′′] = A′′
B′′
C′′
.
But by the hypothesis = ’ ’ ’ , therefore
A′′
B′′
C′′
= ∆A′
B′
C′
∆ABC ≡ ∆A′
B′
C′
.

15
If ∆ ~∆ ’ ’ ’ then (∀) : → ’ ’ ’ and > such that:
|| || = || , ||, , ;
∆ ~∆ ′ ′ ′
‖ ‖
‖ ′ ′‖
=
‖ ‖
‖ ′ ′‖
=
‖ ‖
‖ ′ ′‖
=
A
̂ ≡ A′
̂ ; B
̂ ≡ B′
̂ ; C
̂ ≡ C′
̂
}
‖AB‖ = ‖A′B′‖
‖BC‖ = ‖B′C′‖
‖CA‖ = ‖C′A′‖
.
We construct a function : → ′ ′ ′ such that = ′
, = ′
, = ′;
if | | → | ′ ′| such that || || = || ′ ′||;
if | | → | ′ ′| such that || || = || ′ ′||; – similarity constant.
Let , such that | |, | | ′ | ′ ′| and || || = || ′ ′||
Q′ |A′
C′| and ‖CQ‖ = ‖C′Q′‖ ;
As ‖BC‖ = ‖B′
C′‖ ‖PC‖ = ‖BC‖ − ‖BP‖ = ‖B′
C′‖ − ‖B′
P′‖ =
= ‖B′C′‖ − ‖B′P′‖ = ‖P′C′‖ (2);
C
̂ ≡ C′
̂ (3).
From (1), (2), and (3) ∆PCQ~∆P′
C′
Q′ ‖PQ‖ = ‖P′Q′‖ .
Similar reasoning for , .
We also extend the bijective function previously constructed to the interiors of
the two triangles in the following way:
Let int. and we construct ’ int. ’ ’ ’ such that || || = || ’ ’|| (1).
Let int. → ′ int. ’ ’ ’ such that BAQ
̂ ≡ B’A’Q’
̂ and || || = || ’ ’|| (2).

16
From (1) and (2),
AP
A′P′
=
AQ
A′Q′
= , PAQ
̂ ≡ P′A′Q′
̂ ∆APQ ~ ∆A′
P′
Q′ ‖PQ‖ = ‖P′
Q′‖,
but , [ ], so [ ] ~[ ′ ′ ′].
a. Let | and | ′ ′ be two rays:
Let : | → | ′ ′ such that = ′ and = ′ with || || = || ′ ′|.
The so constructed point ′ is unique and so if ≠ || || ≠ || ||
|| ′ ′|| ≠ || ′ ′|| ′ ≠ ′ and ∀ ′ | ′ ′ a single point | such that
|| || = || ′ ′||.
The constructed function is bijective.
If , | , | | → ′ ′
| ′ ′
such that ‖OP‖ = ‖O′P′‖;‖OQ‖ = ‖O′
Q′‖
‖PQ‖ = ‖OQ‖ − ‖OP‖ = ‖O′
Q′‖ − ‖O′
P′‖ = ‖P′
Q′‖ ∀ P; |OA
the two rays are congruent.
b. Let and ′ be two lines.
Let and ′ ′. We construct a function : → ′ such that = ′ and
| = | ′ ′ and | = | ′ ′ as at the previous point.
It is proved in the same way that is bijective and that || || = || ′ ′|| when
and belong to the same ray.

17
If , belong to different rays:
‖ ‖ = ‖ ′ ′‖
‖ ‖ = ‖ ′ ′‖
} ‖ ‖ = ‖ ‖ + ‖ ‖ = ‖ ′ ′‖ + ‖ ′ ′‖ = ‖ ′ ′‖
and so the two rays are congruent.
We construct a function : → ′ such that = ′, = ′ and a point
∀ → ′ ′ which are considered to be positive.
From the axiom of segment and angle construction that the so constructed
function is bijective, establishing a biunivocal correspondence between the elements
of the two sets.
Let → ′ ′ such that || ′|| = || ||; ̂ ≡ ′ ′ ′
̂ .
As:
‖ ‖ = ‖ ′ ′‖
‖ ‖ = ‖ ′ ′‖
̂ ≡ ′ ′ ′
̂
} ∆ ≡ ′ ′ ′
̂ ‖ ‖ = ‖ ′ ′‖, ∀ , ≡ ′
.

18
: → ′ is an isometry is bijective and ∀ , we have || || =
|| , ||, – bijective – invertible and −
– bijective.
‖ ′ ′‖ = ‖ ; ‖ = ‖ ‖
‖ −
′ ; −
′ ‖ = ‖ −
( ), −
( )‖ = ‖ ‖
}
‖ ′ ′‖ = ‖ −
( ′ ), −
( ′ )‖, ∀ ′
, ′
,
therefore −
: ′ → is an isometry.
We construct a function such that = ′
, = , , … , , and if
| , + |.
The previously constructed function is also extended inside the polygon as
follows: Let int. → ′ int. ′ such that � �+
̂ ≡ ′ ′� ′�+
̂ and ‖ ‖ =
‖ ′ ′ ‖. We connect these points with the vertices of the polygon. It can be easily
proved that the triangles thus obtained are congruent.
We construct the function : [ ] → [ ′] such that
= {
, if
′
, if =
′
, if [ + ] such that
�
̂ ≡ �
′ ′ ′
̂ ∀ = , , … , −
The so constructed function is bijective ∀ , [ ]. It can be proved by the
congruence of the triangles and ′ ′ ′ that || || = || ′ ′||, so [ ] = [ ′]
if two convex polygons are decomposed
in the same number of triangles
respectively congruent,
they are congruent.

19
= … , ; ′
= ′ ′
… , ′
~ ′ > and : → ′ such that ‖ ‖ = ‖ ‖ ∀ , ,
and ′
= .
Taking consecutively the peaks in the role of and , we obtain:
‖ ‖ = ‖ ′ ′‖ ⇒
‖ ‖
‖ ′ ′‖
=
‖ ‖ = ‖ ′ ′‖ ⇒
‖ ‖
‖ ′ ′‖
=
‖ − ‖ = ‖ −
′ ′‖ ⇒
‖ − ‖
‖ −
′ ′‖
=
‖ ‖ = ‖ ′ ′‖ ⇒
‖ ‖
‖ ′ ′‖
=
}
=
‖ ‖
‖ ′ ′‖
=
‖ ‖
‖ ′ ′‖
= =
‖ ‖ + ‖ ‖ + + ‖ − ‖ + ‖ ‖
‖ ′ ′‖ + ‖ ′ ′‖ + + ‖ −
′ ′‖ + ‖ ′ ′‖
= ′
.
�[ ] =
∙
= ; �[ ] = ∙ = = ‖ ‖ ‖ ‖ = = .

20
ℎ = ∙ sin ;
�[ ] =
∙ℎ
is max. when ℎ is max.
max. ℎ = when sin =
⇒ = ⇒ has a right angle at .
‖ ‖ = ‖ ‖ – an isosceles triangle.
(̂) = (̂) = ;
The same way, ( ̂ ) = ( ̂ ) = (̂) = (̂).
‖ ‖ ‖ ‖ = ‖ ‖ = ‖ ‖ = ‖ ‖ is a square.
‖ ‖ = , ‖ ‖ = , ‖ ‖ = √ + =
√
;
‖ ‖ = ‖ ‖ = ‖ ‖ =
√
=
√
;
‖ ‖ =
√
−
√
=
√
;
�[ ] = = �[ ].
�[ ] =
‖ ‖ ∙ ‖ ‖
�[ ] =
‖ ‖ ∙ ‖ ‖
‖ ‖ = ‖ ‖ }
�[ ] = �[ ]

21
�[ ] = �[ ] + �[ ]
�[ ] = �[ ] =
‖ ‖ ∙ ‖ ‖
�[ ] = �[ ] =
‖ ‖ ∙ ‖ ‖
}
�[ ] =
‖ ‖ ‖ ‖ + ‖ ‖
=
‖ ‖ ∙ ℎ
The same way,
�[ ] =
‖ ‖ ∙ ℎ
.
Therefore,
�[ ] = �[ ]
‖ ‖ ∙ ℎ
=
‖ ‖ ∙ ℎ
‖ ‖ = ‖ ‖.
‖ ‖ = ‖ ‖ ;
We draw ⊥ ; ;
‖ ‖ = ‖ ‖ =
ℎ
;
�[ ] =
‖ ‖ + ‖ ‖ ∙ ℎ
−
‖ ‖ ∙ ℎ
−
‖ ‖ ∙ ℎ
=
‖ ‖ + ‖ ‖ ∙ ℎ
= �[ ];
Therefore, [ ] = �[ ] .
�[ ′] is ct. because , , , ′ are fixed points.
Let a line through , and we draw to sides and .
No matter how we draw a line through , �[ ] is formed of: �[ ] +
�[ ] + �[ ].
We have �[ ] constant in all triangles .

22
Let’s analyse:
�[ ′ ] + �[ ] =
‖ ′ ‖ ∙ ℎ
+
‖ ′‖ ∙ ℎ
=
‖ ′ ‖
ℎ +
‖ ‖
‖ ‖
∙ ℎ
=
‖ ′ ‖
∙ (ℎ +
ℎ
ℎ
∙ ℎ ) =
‖ ′ ‖
ℎ
[ ℎ − ℎ + ℎ ℎ ].
∆ is minimal when ℎ = ℎ is in the middle of | |. The construction is
thus: ∆ where || ′. In this case we have | | ≡ | |.
�[ ] =
‖ ‖ ∙ ‖ ′‖
Let the median be | |, and be the centroid of the triangle.
Let ⊥ . �[ ] =
‖ ‖∙‖ ‖
.
′ ⊥
⊥
} ⇒ ′
⇒ ∆ ~∆ ′
⇒
‖ ‖
‖ ′‖
=
‖ ‖
‖ ‖
= ⇒ ‖ ‖ =
‖ ′‖
⇒ �[ ]
=
‖ ‖ ∙
‖ ′‖
=
‖ ‖ ∙ ‖ ′‖
= �[ ].
We prove in the same way that �[ ] = �[ ] = �[ ], so the specific
point is the centroid.

23
Let , such that | |.
We take ′ , ′ .
∆ ′
~∆ ⇒
�[ ′]
�[ ]
= ( )
�[ ′] = �,
‖ ‖
‖ ‖
= , ‖ ‖ =
‖ ‖
√
; ∆ ′~∆
�[ ′]
�[ ]
=
‖ ‖
‖ ‖
�[ ′] = �[ ]
}
‖ ‖
‖ ‖
= ‖ ‖ = √ ‖ ‖ .
‖ ‖ = , ‖ ‖ = ;
�[∆ ′ ] = �[ ′ ′ ] = �[ ′ ] = �[ ] ;
∆ ~∆
�[ ]
�[ ]
=
‖ ‖
‖ ‖
=
�[ ]
�[ ] − �[ ]
=
−
�[ ]
�[ ]
=
−

24
∆ ~∆ ′
�[ ]
�[ ′]
=
‖ ‖
‖ ‖
�[ ]
�[ ′] − �[ ]
=
‖ ‖
�[ ]
�[ ′]
=
‖ ‖ −
�[ ]
�[ ]
=
‖ ‖ −
∆ ′
~∆
�[ ]
�[ ′]
=
‖ ‖
‖ ‖
=
‖ ‖
�[ ]
�[ ′] − �[ ]
=
‖ ‖
�[ ]
�[ ′]
=
‖ ‖ −
�[ ]
�[ ]
=
‖ ‖ −
We divide (1) by (3):
=
‖ ‖ −
−
‖ ‖ − = − ‖ ‖ =
+
.
‖ ‖ = → ‖ ‖ = ; � =
� √
a square inscribed in the circle with radius
= √ = ‖ ‖ √ = = √
‖ ‖ = = √
�[ ] =
‖ ‖ ∙ ‖ ‖ sin
=
√ ∙ √ ∙
√
=
√
�[ ] = �[ ] =
√
=
√
From (1) and (2) �[ ] = � .

25
‖ ‖ = ‖ ‖ ∙ ‖ ′‖
We construct the squares on the hypotenuse and on the leg.
We draw ′ ⊥ .
�[ ] = ‖ ‖
�[ ′ ] = ‖ ‖ ∙ ‖ ′‖ = ‖ ‖ ∙ ‖ ′‖ …
� = �[ ] − �[sect. ]
�[ ] =
√
=
√
= √
�[sect. ] = ( ̂ )
( ̂ ) =
�
( ̂ ) =
�
∙ =
�
�[sect. ] = ∙
�
=
�
� = √ −
�
= √ −
�

26
� = �[sect. ] − �[sect. ] − �[sect. ] − �[sect. ]
= ∙
�
∙ − √ − ∙
�
∙ = ∙
�
− √ −
�
−
�
=
�
−
�
− √
From (1) and (2)
� + � =
�
−
�
−
�
=
�
=
�
.
‖ ‖ = −
�[ , ] = �
�[ , ] = �
�[annulus] = � − � = � −
�[disk diameter‖ ‖] = �‖ ‖ = � −
From (1) and (2) �[annulus] = �[disk diameter].

27
�[ ] =
�[ ′ ′]
�[ ′ ′] = � [ ′ ′] − � [ ′]
We denote (̂ ) = ( ̂ ) = ( ̂ ) =
�
−
� . = − sin
�[ ′ ′] = [� − − sin � − ] = [� − − sin � − ]
� = − sin
�[ ′ ′] = �[ ′ ′] − �[ ′] = � − − sin = − sin
= � − − sin − + sin = � −
�[ ] =
�[ ′ ′]
= � − =
�
−
�[sect. ] = ( ̂ ) =
�
−
From (1) and (2) �[ ] = �[sect. ].
‖ ‖ = ‖ ‖
�[square] = ‖ ‖ = ‖ ‖ = = [ ]
(̂) =
�
�[ ] =
sin
�
=
√
=
√
�[orthogon] = ∙
√
= √

28
�[ ] = �[ ] + �[ ] + �[ ]
ℎ = + +
+ + = ℎ ( is the side of equilateral triangle)
+ + =
√
(because ℎ =
√
).
− given ∆ , , , ℎ − constant
�[ ] =
ℎ
�[ ] = �[ ] + �[ ]
ℎ
= + + =
ℎ
+
ℎ
= + = ,
where = ℎ
and = ℎ
.

29
We draw ′
⊥ ; ′
⊥ ; ′ ⊥ ⇒
′ ′ ′
‖ ‖ = ‖ ‖
} ′
median line in
the trapezoid ′ ′ ‖ ′‖ =
‖ ′‖+‖ ′‖
, �[ ] =
‖ ‖+‖ ′‖
.
First, we construct a quadrilateral with the same area as the given pentagon. We
draw through C a parallel to BD and extend |AB| until it intersects the parallel at M.
�[ ] = �[ ] + �[ ],
�[ ] = �[ ] (have the vertices on a parallel at the base).
Therefore, �[ ] = �[ ].
Then, we consider a triangle with the same area as the quadrilateral .
We draw a parallel to , is an element of the intersection with the same
parallel.
�[ ] = �[ ] + �[ ] = �[ ] + �[ ] = �[ ].

30
‖ ‖ = ‖ ‖
‖ ‖ �[ ] = �[ ]
�[ ] = �[ ] (1)
�[ ] = �[ ] equal bases and the same height;
�[ ] = �[ ]
�[ ] = �[ ] (2)
�[ ] = �[ ] the same base and the vertices on parallel lines at the base;

31
It is proved in the same way that:
rhombus with right angle is a square.
It is proved in the same way that all the peaks of the octagon are elements of the
axis of symmetry of the square, thus the octagon is regular.

32
Consider the square separately.
‖ ‖ = ‖ ‖ = ‖ ‖
‖ ‖? ∆ = ∆ ‖ ‖ = ‖ ‖
It is proved in the same way that ∆ = ∆ ‖ ‖ = ‖ ‖.
Thus is a parallelogram.

33
To determine the angle :
We write
thus we have established the positions of the lines of the locus.
constant for , , , – fixed points.
We must find the geometrical locus of points such that the ratio of the
distances from this point to two concurrent lines to be constant.

34
‖ ‖
‖ ‖
= . Let M' be another point with the same property, namely
‖ ′ ′‖
‖ ′ ′‖
= .
, , ′ collinear the locus is a line that passes through .
When the points are in ∢ we obtain one more line that passes through .
Thus the locus is formed by two concurrent lines through , from which we
eliminate point , because the distances from to both lines are 0 and their ratio is
indefinite.
Vice versa, if points and ′ are on the same line passing through , the ratio
of their distances to lines and is constant.
We show in the same way as in the previous problem that:
‖ ‖
‖ ‖
=
‖ ‖
‖ ‖ + ‖ ‖
=
−
‖ ‖ =
+
,
and the locus of the points which are located at a constant distance from a given
line is a parallel to the respective line, located between the two parallels.
If || || > || || < .

35
Then, if
=
−
=
−
=
−
=
−
,
thus we obtain one more parallel to .
Solution no. 1
We suppose that ABCD is not a parallelogram. Let { } = . We build
such that = and such that = . If a point that verifies
�[ ] + �[ ] = (1), then, because �[ ] = �[ ] and �[ ] = �[ ], it
results that �[ ] + �[ ] = (2).
We obtain that �[ ] = .
On the other hand, the points , are fixed, therefore �[ ] = ′
= const. That
is, �[ ] = − ′
= const.
Because = const., we have , =
( − ′)
= const., which shows that
belongs to a line that is parallel to , taken at the distance
( − ′)
.
Therefore, the locus points are those on the line parallel to , located inside the
quadrilateral . They belong to the segment [ ′ ′] in Fig. 1.
Reciprocally, if [ ′ ′], then �[ ] + �[ ] = �[ ] + �[ ] =
�[ ] = �[ ] + �[ ] = ′
+
∙ ( − ′)
∙
= .

36
In conclusion, the locus of points inside the quadrilateral which occurs
for relation (1) where is a positive constant smaller than = �[ ] is a line
segment.
If is a trapeze having and as bases, then we reconstruct the
reasoning as = { } and �[ ] + �[ ] = − = const.
If is a parallelogram, one shows without difficulty that the locus is a
segment parallel to .
Solution no. 2 (Ion Patrascu)
We prove that the locus of points which verify the relationship �[ ] +
�[ ] = (1) from inside the convex quadrilateral of area ( ⊂ ) is a line
segment.
Let’s suppose that = { }, see Fig. 2. There is a point of the locus which
belongs to the line . Therefore, we have ; = . Also, there is the point
such that ; = .
Now, we prove that the points from inside the quadrilateral that are on
the segment [ ] belong to the locus.
Let int[ ] [ ]. We denote and the projections of on and
respectively. Also, let be the projection of on and the projection of
on . The triangles and are alike, which means that
= ,

37
and the triangles and are alike, which means that
= .
By adding member by member the relations (2) and (3), we obtain
+ =
+
= .
Substituting in (4), = and = , we get ∙ + ∙ = , that
is �[ ] + �[ ] = .
We prove now by reductio ad absurdum that there is no point inside the
quadrilateral that is not situated on the segment [ ], built as shown, to
verify the relation (1).
Let a point ′
inside the quadrilateral that verifies the relation (1), ′
[ ]. We build ′
, ′
, where and are situated on [ ], see Fig. 3.
We denote ′
, , the projections of , , on and ′
, , the
projections of the same points on .
We have the relations:
′ ′
∙ + ′ ′
∙ = ,
∙ + ∙ = .
Because ′ ′
= and ′ ′
= , substituting in (5), we get:
∙ + ∙ = .
From (6) and (7), we get that = , which drives us to , false!

38
Problems in Geometry and Trigonometry
39. Find the locus of the points such that the sum of the distances to two
concurrent lines to be constant and equal to .
40. Show that in any triangle we have:
a. + = ; b. + = − .
41. Show that among the angles of the triangle we have:
a. − =
−
;
b. + + = + + .
42. Using the law of cosines prove that = + − , where is the
length of the median corresponding to the side of length.
43. Show that the triangle where
+
= cot is right-angled.
44. Show that, if in the triangle we have cot + cot = cot ⇒ +
= .
45. Determine the unknown elements of the triangle , given:
a. , and ;
b. + = , and ;
c. , ; − = .

39
46. Show that in any triangle we have tan
−
tan =
−
+
(tangents
theorem).
47. In triangle it is given ̂ = ° and = + √ . Find tan
−
and angles
and .
48. In a convex quadrilateral , there are given ‖ ‖ = (√ − √ ), ‖ ‖ =
, ‖ ‖ = , = arccos , and =
�
+ arccos . The other angles of the
quadrilateral and ‖ ‖ are required.
49. Find the area of ∆ when:
a. = , = arcsin , = arcsin ;
b. = , ̂ °, ̂ °;
c. = , = , = ;
d. ̂ °, = , = .
50. How many distinct triangles from the point of view of symmetry are there
such that = , = , = ?
51. Find the area of ∆ if = √ , ̂ °, + = + √ .
52. Find the area of the quadrilateral from problem 48.

40
53. If is the area of the regular polygon with sides, find:
, , , , , in relation to , the radius of the circle inscribed in the
polygon.
54. Find the area of the regular polygon … inscribed in the circle with
radius , knowing that: ‖ ‖
= ‖ ‖
+ ‖ ‖
.
55. Prove that in any triangle ABC we have:
a. = − tan ;
b. = − tan ;
c. = cos cos cos ;
d. − = cos cos cos ;
e. = sin + cos sin sin ;
f. ℎ = sin sin .
56. If is the center of the circle inscribed in triangle show that ‖ ‖ =
sin sin .
57. Prove the law of sine using the analytic method.
58. Using the law of sine, show that in a triangle the larger side lies opposite
to the larger angle.
a.
cos − cos
cos − cos
+ cos = , ≠ ;

41
b.
sin − sin
+ cos − cos
=
−
+
;
c. + cos + cos + = cos cos .
60. In a triangle , °, ‖ ‖ = , ‖ ‖ =
√
. Show that tan = .
61. Let ′, ′, ′ be tangent points of the circle inscribed in a triangle with
its sides. Show that
�[ ′ ′ ′]
�[ ]
=
�
.
62. Show that in any triangle sin
√
.
63. Solve the triangle , knowing its elements , and area .
64. Solve the triangle , knowing = , arc cos , and the corresponding
median for side , = √ √ .
65. Find the angles of the triangle , knowing that − =
�
and = ,
where and are the radii of the circles circumscribed and inscribed in the
triangle.

42
Solutions
Let and be the two concurrent lines. We draw 2 parallel lines to located
on its both sides at distance . These intersect on at and , which will be
points of the locus to be found, because the sum of the distances , +
, = + verifies the condition from the statement.
We draw two parallel lines with located at distance from it, which cut in
and , which are as well points of the locus to be found. The equidistant parallel
lines determine on congruent segments
| | ≡ | |
| | ≡ | |
, in the same way
is a parallelogram.
∆ ,
‖ ′‖ = ,
‖ ′‖ = ,
} ‖ ′‖ = ‖ ′‖
∆ is isosceles.
|| || = || || is a rectangle. Any point we take on the sides of this
rectangle, we have || , || + || , || = , using the propriety according to which
the sum of the distances from a point on the base of an isosceles triangle at the
sides is constant and equal to the height that starts from one vertex of the base,
namely . Thus the desired locus is rectangle .

43
= + − cos ;
= + − cos = + −
+ −
= + − − +
= + − = + − .

44
Using the sine theorem, = sin .
−
= − − = or − =
= + = =
or or or
+ = = =
By substitution:

45
a. Using the law of sine,
b.
c.
Therefore,
We solve the system, and find and . Then we find = and = − .
= = =
= sin
= sin
;
−
+
=
sin − sin
sin + sin
=
sin − sin
sin + sin
=
sin
−
cos
+
sin
+
cos
−
= tan
− sin
cos
= tan
−
tan .

46
Using tangents’ theorem,
So
In ∆ we have

47
In ∆ ,
In ∆ we apply sine’s theorem:
Or we find ̂ and we add it to
�
.

48
sin =
− +√
, because < and sin > .

49

50
At problem 9 we’ve found that
With Heron’s formula, we find the area of each triangle and we add them up.
The formula for the area of a regular polygon:

51
In ∆ :
In ∆ :
In ∆ :
Substituting (1), (2), (3) in the given relation:
or
which is impossible.
=
complete circle
( ̂ )
=
�
�
= .
Thus the polygon has 7 sides.

52

53
We apply the law of sine in ∆ :
The law of sine applied in ∆ :
In ∆ ′: sin − =
‖ ′‖
‖ ′‖ = sin ; cos − = cos .
So the coordinates of are − cos , sin .
The center of the inscribed circle is at the intersection of the perpendicular lines
drawn through the midpoints of sides and .
The equation of the line :

54
If we redo the calculus for the same draw, we have the following result:
cos , sin .
using the law of cosine.
sin
=
sin
=
sin
= .
We suppose that > . Let’s prove that > .
sin
=
sin
⇒ =
sin
sin
> ⇒ >
}
sin
sin
> , , , � sin > sin > sin
sin − sin > sin
−
cos
+
>
+
=
−
= − .

55
cos
+
= cos − = sin > , therefore
−
> > ;
(−
�
<
−
<
�
).
b. We transform the product into a sum:
From (1) and (2)
We consider the last two terms:

56
We apply the law of cosines in triangle ABC:
‖ ‖ = ‖ ‖ = ‖ ‖ =
′ ⊥
′ ⊥
} ⇒ ′ ′ ′ inscribable quadrilateral
′ ′ = − ̂ sin( ′ ′
̂ ) = sin ̂
Similarly, ′ ′
̂ = sin and ′ ′
̂ = sin .

57
In the same way,
< ⇒< <
�
⇒ sin > ;
sin = √
− cos sin =
− cos
cos =
+ −
}
sin =
−
− +
=
− −
sin
√
.
= � − +
=
, , are known
} We find .
= = . In the same way, we find .

58
+ =
=
} {
=
=
or {
=
=
We find .
We find the sum.
Or
We already know that
= sin sin sin
We write sin = . We have

59
From this system we find and .

60
Other Problems in Geometry
and Trigonometry (10th
grade)
66. Show that a convex polygon can’t have more than three acute angles.
67. Let be a triangle. Find the locus of points , for which
�[ ] = �[ ].
68. A convex quadrilateral is given. Find the locus of points
. , for which �[ ] = �[ ].
69. Determine a line , parallel to the bases of a trapezoid (
| |, | |) such that the difference of the areas of [ ] and [ ]
to be equal to a given number.
70. On the sides of ∆ we take the points , , such that = = = .
Find the ratio of the areas of triangles and .
71. Consider the equilateral triangle and the disk [ , ], where is the
orthocenter of the triangle and = ‖ ‖. Determine the area [ ] −
[ , ].

61
a. + cos cos − =
+
;
b. + = tan = ;
c.
+
� =
�
+
+
;
d. = cot + cot + cot ;
e. cot + cot + cot =
�
.
73. If is the orthocenter of triangle , show that:
a. ‖ ‖ = cos ;
b. ‖ ‖ + ‖ ‖ + ‖ ‖ = .
74. If is the orthocenter of the circumscribed circle of triangle and is
the center of the inscribed circle, show that ‖ ‖ = – .
75. Show that in any triangle we have: cos
− �
.
76. Find +
��
knowing that +
�
= sin .
77. Solve the equation: + − − = .
78. Prove that if < then | + + | .

62
79. One gives the lines and ′. Show that through each point in the space
passes a perpendicular line to and ′.
80. There are given the lines and ′, which are not in the same plane, and
the points , ′. Find the locus of points for which pr = and
pr ′ = .
81. Find the locus of the points inside a trihedral angle ̂ equally distant
from the edges of , , .
82. Construct a line which intersects two given lines and which is perpendicular
to another given line.
83. One gives the points and located on the same side of a plane; find in
this plane the point for which the sum of its distances to and is
minimal.
84. Through a line draw a plane onto which the projections of two lines to be
parallel.
85. Consider a tetrahedron [ ] and centroids , , of triangles
, , .
a. Show that ;
b. Find the ratio
�[ ]
�[ ]
.

63
86. Consider a cube [ ′ ′ ′ ′]. The point is projected onto ′ , ′ , ′
respectively in , , . Show that:
a. ′
⊥ ;
b. ⊥ , ⊥ ;
c. is an inscribable quadrilateral.
87. Consider the right triangles and (̂) = ( ̂ = )
located on perpendicular planes and , being midpoints of segments
[ ], [ ]. Show that ⊥ .
88. Prove that the bisector half-plane of a dihedral angle inside a tetrahedron
divides the opposite edge in proportional segments with the areas of the
adjacent faces.
89. Let be a vertex of a regular tetrahedron and , two points on its
surface. Show that ̂ .
90. Show that the sum of the measures of the dihedral angles of a tetrahedron
is bigger than 3“0°.
91. Consider lines , contained in a plane and a line which intersects
plane at point . A variable line, included in and passing through all
, respectively at . Find the locus of the intersection . In
which case is the locus an empty set?

64
92. A plane intersects sides [ ], [ ], [ ], [ ] of a tetrahedron [ ] at
points , , , . Prove that ‖ ‖ ∙ ‖ ‖ ∙ ‖ ‖ ∙ ‖ ‖ = ‖ ‖ ∙ ‖ ‖ ∙
‖ ‖ ∙ ‖ ‖.
93. From a point located outside a plane , we draw the perpendicular line
, , and we take , . Let , be the orthocenters of triangles
, ; and heights in triangle ; and height in triangle
. Show that:
a. ⊥ ;
b. ‖ ‖ ∙ ‖ ‖ ∙ ‖ ‖ = .
94. Being given a tetrahedron [ ] where ⊥ and ⊥ , show that:
a. ‖ ‖ + ‖ ‖ = ‖ ‖ + ‖ ‖ = ‖ ‖ + ‖ ‖ ;
b. The midpoints of the 6 edges are located on a sphere.
95. It is given a triangular prism [ ′ ′ ′] which has square lateral faces. Let
be a mobile point [ ′], the projection of onto ′ and ʺ the
midpoint of [ ′ ʺ]. Show that ′ and ʺ intersect in a point and find
the locus of .
96. We have the tetrahedron [ ] and let be the centroid of triangle
. Show that if then �[ ] = �[ ] = �[ ].
97. Consider point the interior of a trirectangular tetrahedron with its
vertex in . Draw through a plane which intersects the edges of the

65
respective tetrahedron in points , , so that is the orthocenter of
∆ .
98. A pile of sand has as bases two rectangles located in parallel planes and
trapezoid side faces. Find the volume of the pile, knowing the dimensions
′, ′ of the small base, , of the larger base, and ℎ the distance between
the two bases.
99. A pyramid frustum is given, with its height ℎ and the areas of the bases
and . Unite any point � of the larger base with the vertices , , ′, ′ of a
side face. Show that �[ ′ ′ ] =
√
√
�[ ′].
100. A triangular prism is circumscribed to a circle of radius . Find the area
and the volume of the prism.
101. A right triangle, with its legs and and the hypotenuse , revolves by
turns around the hypotenuse and the two legs, , , ; , , being the
volumes, respectively the lateral areas of the three formed shapes, show
that:
a.
�
=
�
=
�
;
b. + =
+
.
102. A factory chimney has the shape of a cone frustum and 10m height, the
bases of the cone frustum have external lengths of 3,14m and 1,57m, and
the wall is 18cm thick. Calculate the volume of the chimney.

66
103. A regular pyramid, with its base a square and the angle from the peak of
a side face of measure is inscribed in a sphere of radius . Find:
a. the volume of the inscribed pyramid;
b. the lateral and total area of the pyramid;
c. the value when the height of the pyramid is equal to the radius of
the sphere.

67
Solutions
Let , … the vertices of the convex polygon. Let’s assume that it has four
acute angles. The vertices of these angles form a convex quadrilateral .
Due to the fact that the polygon is convex, the segments | |, | |, | |,
| | are inside the initial polygon. We find that the angles of the quadrilateral are
acute, which is absurd, because their sum is 3“0°.
Another solution: We assume that is a convex polygon with all its
angles acute the sum of the external angles is bigger than 3“0°, which is absurd
(the sum of the measures of the external angles of a convex polygon is 3“0°).

68
Let | ′| be the median from and ⊥ ′
, ⊥ ′.
∆ ′ ≡ ′ because:
{
̂ ≡ ̂ alternate interior
′
̂ ≡ ′
̂ vertical angles
′
≡ ′
|| || = || || and by its construction ⊥ ′, ⊥ ′.
The desired locus is median | ′|. Indeed, for any | ′| we have �[ ] =
�[ ], because triangles and have a common side | | and its
corresponding height equal || || = || ||.
Vice-versa. If �[ ] = �[ ′ ], let’s prove that | ′|.
Indeed: �[ ] �[ ] ⇒ , = , , because | | is a common
side, , = || || and , = || || and both are perpendicular to
is a parallelogram, the points , , are collinear ( , the feet of the
perpendicular lines from and to ).
In parallelogram we have | | and | | diagonals passes through
the middle of | |, so | ′|, the median from .
Let be the midpoint of diagonal | | ‖ ‖ = ‖ ‖.
�[ ] = �[ ] (1)
Because {
‖ ‖ = ‖ ‖
‖ ′‖ common height
�[ ] = �[ ] (2)

69
the same reasons; we add up (1) and (2)
�[ ] = �[ ] (3),
so is a point of the desired locus.
We construct through a parallel to until it cuts sides | | and | | at
respectively . The desired locus is | |.
Indeed ∀ | | we have:
�[ ] = �[ ] because and belongs to a parallel to .
, a parallel to .
So
�[ ] = �[ ]
and �[ ] = �[ ]
and from
} �[ ] = �[ ].
Vice-versa: If �[ ] = �[ ], let’s prove that parallel line through to
. Indeed:
�[ ] = �[ ]
and because �[ ] + �[ ] = �[ ]
} �[ ] = �[ ] =
�[ ]
(2).
So, from (1) and (2) �[ ] = �[ ] �[ ] + �[ ] = �[ ] +
�[ ] �[ ] and are on a parallel to .
We write || || = and || || = , || || = .

70
We subtract (2) from (3)
We subtract (2) from (4)
From the relation (3), by writing [ ] − �[ ] = −
.
We substitute this in the relation of ² and we obtain:
and taking into consideration that || || = || || + , we have
so we have the position of point on the segment | | (but it was sufficient to
find the distance || ||).
We remark from its construction that || || , more than that, they are
equidistant parallel lines. Similarly, , , and , , are also equidistant
parallel lines.
from the hypothesis

71
We write �[ ] = .
Based on the following properties:
 two triangles have equal areas if they have equal bases and the same
height;
 two triangles have equal areas if they have the same base and the third
peak on a parallel line to the base,
we have:
⇒
So
�[ ]
�[ ]
= = .
(If necessary the areas can be arranged).
‖ ‖ =
√
( ′ median)
In ∆ ′:
So ( ̂ ) =
�
.
We mark with � the disk surface bordered by a side of the triangle outside the
triangle.

72
�[�] = �[circle sector ] − �[ ]
=
�
∙
−
∙
sin =
�
∙
−
√
∙
= ∙
�
−
√
.
If through the disk area we subtract three times �[�], we will find the area of the
disk fraction from the interior of . So the area of the disk surface inside is:
The desired area is obtained by subtracting the calculated area form �[ ].
So:
a. + cos ∙ cos − =
+
b. We prove that tan =
+ −
.

73
All terms reduce.
d. Let’s prove that cot + cot + cot =
�
.
Indeed
We now have to prove that:

74
a. In triangle ′: ‖ ′‖ = cos
In triangle ′:
We used:
Using the power of point in relation to circle ,
Taking into consideration (1), we have ‖ ‖ ∙ ‖ ‖ = − ‖ ‖².
We now find the distances ||IA|| and ||ID||
In triangle ∆ ,
We also find ‖ ‖: ( ̂ ) = ( ̂ ) have the same measure, more exactly:

75
In ∆ according to the law of sine, we have:
So taking into consideration (3),
Returning to the relation ‖ ‖ ∙ ‖ ‖ = − ‖ ‖². with (2) and (4) we have:
Note. We will have to show that
Indeed:
(by Heron’s formula).

76
So:
We calculate for and :
so +
��
takes the same value for and for and it is enough if we
calculate it for .
Analogously:

77
(we substitute − with ² at denominator)
We construct ⊥ and . The so constructed plane is unique. Similarly we
construct ⊥ ′ and ⊥ , = ∋ .
From
⊥ ⇒ ⊥
⊥ ′ ⇒ ′ ⊥
} is a line which passes through and is perpendicular
to and ′. The line is unique, because and constructed as above are unique.
We construct plane such that and ⊥ . We construct plane such that
and ′ ⊥ .

78
The so constructed planes and are unique.
Let = ⊂ so (∀) has the property pr = .
⊂ ∀ has the property ′ = .
Vice-versa. If there is a point in space such that pr = and ′ =
and ( and previously constructed).
Let , , such that ‖ ‖ = ‖ ‖ = ‖ ‖. Triangles , ,
are isosceles. The mediator planes of segments ‖ ‖, ‖ ‖, ‖ ‖ pass through
and ′ (the center of the circumscribed circle of triangle ). Ray | ′| is the
desired locus.
Indeed (∀) | ′| mediator plane of segments | |, | | and | |
is equally distant from , and .
Vice-versa: (∀) with the property: , = , = , mediator
plan, mediator planes of segments | |, | | and | | the intersection of
these planes | ′|.

79
Let , , be the 3 lines in space.
I. We assume ⊥ and ⊥ . Let be a plane such that:
The construction is possible because ⊥ and ⊥ . Line meets on and it is
perpendicular to , because ⊂ and ⊥ .
II. If ⊥ or ⊥ , the construction is not always possible, only if plane ,
is perpendicular to .
III. If ⊥ and ⊥ , we construct plane ⊥ so that ⊂ and ⊂ ≠ . Any
point on line a connected with point is a desired line.
We construct ′ the symmetrical point of in relation to . ′ and are on
different half-spaces, | ′ | = .

80
is the desired point, because || || + || || = || ′|| + || || is minimal when
| ′ |, thus the desired point is = | ′ | .
Let , , be the 3 given lines and through we construct a plane in which and
to be projected after parallel lines.
Let be an arbitrary point on . Through we construct line ′|| . It results from
the figure || , = , ′ .
Let such that ⊂ and ⊥ .
Lines and ′ are projected onto after the same line . Line is projected onto
after and .
If ,
absurd because || || ′ .
is the centroid in ∆
| |
| |
= (1)

81
| |
| |
= (2)
| |
| |
= (3)
From 1 and 2,
and from 2 and 3
because:
So
⊥ ′ from the hypothesis ′ ′ ′ ′ cube (1).
midpoint of segment | ′|
′ isosceles and ⊥ ′
midpoint of | ′ |
} | | mid-side in ∆ ′ (2)
From (1) and (2) ′ ′ ⊥ (3)

82
From ∆ ′:
From ∆ ′:
Similarly
In ∆ ′:
and
′ right with ′ = because
From (4) and (3) ′ ⊥ .
As
′ ⊥
′
⊥ by construction
} coplanar quadrilateral with
opposite angles and right inscribable quadrilateral.
The conclusion is true only if || || = || || that is = .
⊥ if

83
bisector plane
( bisector half-plane)
In triangle ′:
But

84
From 1, 2, 3
�[ ]
�[ ]
=
‖ ‖
‖ ‖
q.e.d.
Because the tetrahedron is regular = … =
we increase the denominator
If one of the points or is on face the problem is explicit.
We consider tetrahedron , and prove that the sum of the measures of the
dihedral angles of this trihedron is bigger than °. Indeed: let ′ be the internal
bisector of trihedron ( ′ the intersection of the bisector planes of the 3
dihedral angles) of the trihedron in , , .

85
The size of each dihedron with edges , , is bigger than the size of the
corresponding angles of , the sum of the measures of the dihedral angles of
trihedron is bigger than °.
Let , be a plane ⊥ to at ; ⊥ , ⊥ , but | and | are on the same
half-space in relation to ⇒ ̂ < ̂ .
In tetrahedron , let , , , , and be the 6 dihedral angles formed
by the faces of the tetrahedron.
according to the inequality previously established.
We mark with a the intersection of planes , and , . So
Let be a variable line that passes through and contained in , which cuts
and at respectively . We have: ⊂ , , = and
intersect because they are contained in the plane determined by ( , ).
Thus , and , , , so describes line a the intersection of
planes , and , .
Vice-versa: let .
In the plane , : = ′
In the plane , : = ′

86
Lines ′ ′ and are coplanar (both are on plane , , ). But because ′ ′ ⊂
and has only point in common with ′ ′ = .
So ′ ′ passes through . If planes , and , are parallel, the locus is the
empty set.
Remember the theorem: If a plane intersects two planes and such that
�|| || . If plane || we have:
If || we have:
relation .
Solution:
Let ′, ′, ′, ′ the projections of points , , , onto plane .
For ex. points ′, , ′ are collinear on plane because they are on the
projection of line onto this plane.
Similarly we obtain:

87
By multiplying the 4 relations,
relation from .
– Midpoint of | |.
(1) ⊥ (hypothesis)
From 1 and 2
height in ∆ a
From 3 and 4 ⊥ ⊥ height in ∆ b
From a and b orthocenter ∆ . Let be the diametrical opposite point to
in circle diameter ̂ = but ⊥ . Similarly
|| , so parallelogram, we have:

88
similarly
diametrical opposite to B
but
by substituting above, we have:
Let , , , , , midpoints of the edges of the quadrilateral because:
|| || (median lines), || || (median lines), but ⊥ ⇒
rectangle | | | | = { }.
Similarly rectangle with | | common diagonal with a, the first rectangle,
so the “ points are equally distant from the midpoint of diagonals in the two
rectangles the 6 points are on a sphere.
arbitrary point on | ′|
ʺ midpoint of segment [ ′ ′]
When = ′, point is in the position ′.

89
When = , point is in the position { } = [ ′ ] [ ′′]
( ′ ′′ rectangle, so is the intersection of the diagonals of the rectangle)
[The locus is [ ′ ]].
Let be arbitrary point | ′|.
because:
By the way it was constructed
∀ plane that contains is perpendicular to ′ ′′ , particularly to ′ ⊥
′ ′ .
because ′, ′ ′′
from this reason ′, ′ ′ .
From 1 and 2
Let
So ∀ | ′ |
and we have
Vice-versa. Let arbitrary point, | ′ | and
In plane
In plane
Indeed: ′ ′′ || ′ thus any plane which passes through ′ ′′ will intersect
′ after a parallel line to ′ ′′. Deci || ′ ′′ or || as ′
⊥ ′ ′′ .
We’ve proved

90
and
we have
describes | ′ | and vice-versa,
there is | ′ | and | ′ | such that
and is the intersection of the diagonals of the quadrilateral ′ ′′.
known result
From 1 and 3
From the hypothesis:
⊥ ⊥ ⊥
We assume the problem is solved.
Let be the orthocenter of triangle .

91
But ′ ⊥
⊥ ′
⊥ ′
} ⊥ ⊥ (1)
⊥ ⊥ , but ′ ⊥
⊥ ′
⊂ ′
} ⊥ ⊥ (2)
From (1) and (2) ⊥
So the plane that needs to be drawn must be perpendicular to at .
′ ⊥ , ′ ⊥
‖ ‖ =
− ′
, ‖ ‖ =
− ′
�[ ′] =
− ′
∙
− ′
∙
ℎ
�[ ′ ′] =
�[ ′] ∙ ‖ ′ ′‖
=
− ′
∙
ℎ
∙ ′
�[ ′ ′ ′ ′ ] =
+ ′ ℎ
∙ ′
�[ ′ ′ ′ ′] = [ ∙
− ′
∙
− ′
∙
ℎ
+
− ′
∙
ℎ
∙ ′] +
+ ′ ℎ
∙ ′
=
ℎ
− ′
− ′ ′
+ ′
− ′ ′
+ ′
+ ′ ′
=
ℎ
[ + ′ ′
+
+ ′ + ′ ].

92
�[ ′] =
∙ ℎ
�[ ′ ′ ] = �[ ′ ′ ′] − �[ ′ ] − �[ ′ ′ ′ ] =
ℎ
( + + √ ) −
ℎ
−
ℎ
=
ℎ
√ .
So:
�[ ′ ′ ]
�[ ′]
=
ℎ
√ ∙ √
ℎ
∙
=
√
√
�[ ′ ′ ] =
√
√
∙ �[ ′]
For the relation above, determine the formula of the volume of the pyramid
frustum.
′ = ℎ =
Let = ‖ ‖ ‖ ‖ =
√
‖ ‖ =
√
Figure rectangle ‖ ‖ = ‖ ‖ √
= = √

93
So, the lateral area is = ∙ √ ∙ = √ .
�[ ′ ′ ′] = �[ ] ∙ = √ ∙
√ √
∙ = √ .
The total area:
� = + �[ ] = √ + ∙ √ = √
Let and be the volume, respectively the area obtained revolving around .
and be the volume, respectively the area obtained after revolving around .
and be the volume, respectively the area obtained after revolving around c.
So:
=
� ∙ ‖ ‖ + ‖ ‖
=
� ∙ ∙
= � ∙ ∙ + � ∙ ∙ = � ∙ ∙ +
=
�
=
�
=
�
= � ∙ ∙
=
�
=
�
=
�
= � ∙ ∙
Therefore:
= +
�
=
�
+
�
+ =
+
+ =
� +
� +
+
∙
=
But ∙ = ∙
+
= , ‖ ‖ = .

94
= ‖ ‖ =
= ‖ ′ ‖ =
� = , = ,
� = , = ,
‖ ‖ = = ,
‖ ′ ‖ =
‖ ‖ = √ + , = ,
‖ ′ ‖ =
‖ ′‖ ∙ ‖ ′ ‖
‖ ‖
=
∙ ,
,
≈ ,
‖ ‖
‖ ′ ‖
=
‖ ‖
‖ ′‖
,
,
=
‖ ‖
,
‖ ‖ = ,
‖ ′ ‖ = ′
= , − , = ,
‖ ‖ = ′
= , − , = ,
=
�
+ +
=
�
, + , + , ∙ , − , − , − , ∙ ,
=
�
, − , = , �
‖ ‖ =
sin
cos
In ∆ : ‖ ‖ = � .

95
In ∆ ′: ‖ ′‖ = � − .
‖ ′‖ =
√cos
sin
In ∆ ′: ‖ ′‖ =
√ �
� − .
= +
√cos
sin
− +
cos
sin
−
√cos
sin
=
√cos sin
sin cos + cos
= √cos ∙ sin
=
cos
∙ sin
=
cos
sin
= cos sin ∙
cos
sin
= cos sin = sin
� = + = sin + cos sin
‖ ′‖ = =
√cos
sin
= √cos sin ∙
√cos
sin
cos = = .

96
Various Problems
104. Determine the set of points in the plane, with affine coordinates that
satisfy:
a. | | = ;
b. � < arg
�
; ≠ ;
c. arg >
�
, ≠ ;
d. | + | .
105. Prove that the roots of the unit are equal to the power of the particular
root .
106. Knowing that complex number verifies the equation = , show that
numbers 2, − and verify this equation.
Application: Find − and deduct the roots of order 4 of the number
− + .
107. Show that if natural numbers and are coprime, then the equations
− = and − = have a single common root.
108. Solve the following binomial equation: − + + = .
109. Solve the equations:

97
110. Solve the equation ̅ = −
, , where ̅ the conjugate of .
111. The midpoints of the sides of a quadrilateral are the vertices of a
parallelogram.
112. Let and two parallelograms and the midpoints of
segments [ ], { , , , }. Show that is a parallelogram or a
degenerate parallelogram.
113. Let the function : → , = + ; , , , ≠ . If and
are of affixes and , and ′
and ′
are of affixes , , show
that ‖ ′ ′ ‖ = | | ∙ ‖ ‖. We have ‖ ′ ′ ‖ = ‖ ‖ ⇔ | | = .
114. Prove that the function z → z
̅, z C defines an isometry.
115. Let M M be of affixes , ≠ and z = αz . Show that rays |OM , |OM
coincide (respectively are opposed) α > (respectively α < ).

98
116. Consider the points M M M of affixes and ≠ . Show that:
a. M |M M
−
−
> ;
b. M M M
−
−
R .
117. Prove Pompeiu’s theorem. If the point from the plane of the equilateral
triangle the circumscribed circle ∆  there exists a
triangle having sides of length ‖ ‖, ‖ ‖, ‖ ‖.

99
Solutions
a.
| | =
| | = √ +
} + = , so the desired set is the circle , .
b. � < arg
�
.
The desired set is given by all the points of quadrant III, to which ray | is added,
so all the points with < , < .
c.
arg >
�
, ≠
arg [ , �]
}
�
< arg < �
The desired set is that of the internal points of the angle with its sides positive
semi-axis and ray | .
d. | + | ; = + , its geometric image .
where ′
, − .
Thus, the desired set is the disk centered at ,−
′
and radius 2.

100
Let the equation = . If = ( is the solution) then: − = − = ∙ =
, so – is also a solution.
is the solution;
− is the solution;
is the solution of the equation = − + .
The solutions of this equation are:
but based on the first part, if − − is a root, then
are solutions of the given equation.
If there exist and ′ with = ′, then

101
because , = . Because ′
< , < , we have ′ = , = .
Thus the common root is .
As:

102
From:
positive
The given equation becomes
We find the sum of the abscissa of the opposite points:
a parallelogram.
In the quadrilateral by connecting the midpoints we obtain the
parallelogram ′ ′′
, with its diagonals intersecting at , the midpoint of | ′ ′′|
and thus | | ≡ | |. (1)

103
In the quadrilateral by connecting the midpoints of the sides we obtain
the parallelogram ′ ′′
with its diagonals intersecting in , the midpoint of
| ′ ′′| and thus | | ≡ | |. (2)
From (1) and (2) a parallelogram.
If:
If:
Let and be of affixes and . Their images through the given function
′
and ′
with affixes ̅ and ̅ , so

104
From (1) and (2) ‖ ‖ = ‖ ′ ′ ‖ or ‖ ′ ′ ‖ = | ̅ ̅ | = |√ − | =
| − | = ‖ ‖.
So : → , = ̅ defines an isometry because it preserves the distance
between the points.
We know that the argument = arg + arg � − �, where = or = .
Because arg = arg , arg = arg + arg � − �.
a. We assume that
Vice versa,
b. Let | and | be opposed arg = arg + �
to the negative ray | ′
< . Vice versa,
= or =
arg = arg + �
or
arg = arg − �
} | and | are opposed.

105
If n and ′ are the geometric images of complex numbers and ′, then the image
of the difference – ′ is constructed on |OM | and | ′ | as sides.
We assume that |
We construct the geometric image of – . It is the fourth vertex of the
parallelogram . The geometric image of – is , the fourth vertex of the
parallelogram .
collinear
} , , collinear
Vice versa, we assume that
If and the opposite ray to , then – = ( – ) with < .
We repeat the reasoning from the previous point for the same case.

106
Thus, when + we obtain for the respective ratio positive,
negative or having = , so
−
−
R.
The images of the roots of order 3 of the unit are the peaks of the equilateral
triangle.
But ɛ = ɛ , so if we write ɛ = , then ɛ = ɛ .
Thus , , .
We use the equality:
adequate ∀ ℂ.
But
Therefore,
By substitution:
but

107
thus
Therefore ‖ ‖, ‖ ‖, ‖ ‖ sides of a ∆.
Then we use ‖ | − | ‖ | − | and obtain the other inequality.

108
Problems in Spatial Geometry
118. Show that if a line is not contained in plane , then is or it is
formed of a single point.
119. Show that (∀) , ( ) at least one point which is not situated in .
120. The same; there are two lines with no point in common.
121. Show that if there is a line ( ) at least two planes that contain line .
122. Consider lines , ′, ′′, such that, taken two by two, to intersect. Show
that, in this case, the 3 lines have a common point and are located on the
same plane.
123. Let , , be three non-collinear points and a point located on the
plane . Show that:
a. The points , , are not collinear, and neither are , , ; , , .
b. The intersection of planes , , is formed of a single
point.

109
124. Using the notes from the previous exercise, take the points , , distinct
from , , , , such that , , . Let = { },
= { }, = { }. Show that , , are collinear (T. Desarques).
125. Consider the lines and ′ which are not located on the same plane and
the distinct points , , and , ′. How many planes can we draw
such that each of them contains 3 non-collinear points of the given points?
Generalization.
126. Show that there exist infinite planes that contain a given line .
127. Consider points , , , which are not located on the same plane.
a. How many of the lines , , , , , can be intersected by a
line that doesn’t pass through , , , ?
b. Or by a plane that doesn’t pass through , , , ?
128. The points and are given, , . Construct a point at an
equal distance from and , that also to plan .
129. Determine the intersection of three distinct planes , , .
130. Given: plane , lines , and points , . Find a point
such that the lines , intersect and .

110
131. There are given the plane , the line , the points , , and
. Let and ′, ′ the points of intersection of the lines ,
with plane (if they exist). Determine the point such that the points
, ′, ′ to be collinear.
132. If points and of an open half-space �, then [ ] ⊂ �. The property is
as well adherent for a closed half-space.
133. If point is not situated on plane and then | ⊂ | .
134. Show that the intersection of a line with a half-space is either line or
a ray or an empty set.
135. Show that if a plane and the margin of a half-space � are secant
planes, then the intersection � is a half-plane.
136. The intersection of a plane with a half-space is either the plane or a
half-plane, or an empty set.
137. Let , , , four non coplanar points and a plane that doesn’t pass
through one of the given points, but it passes trough a point of the line
| |. How many of the segments | |, | |, | |, | |, | |, | | can be
intersected by plane ?

111
138. Let be a line and , two planes such that = and = .
Show that if and , then ⊂ | and ⊂ | .
139. Let | and | be two half-spaces such that ≠ and | ⊂ | or
| | = . Show that = .
140. Show that the intersection of a dihedral angle with a plane can be: a
right angle, the union of two lines, a line, an empty set or a closed half-
plane and cannot be any other type of set.
141. Let be the edge of a proper dihedron ∠ ′ ′
, ′
– , ′
– and
int. ∠ ′ ′
. Show that:
a. int. ∠ ′ ′
= | ;
b. If , int. ∠ = int. ′ ′
.
142. Consider the notes from the previous problem. Show that:
a. The points and are on different sides of the plane ;
b. The segment | | and the half-plane | have a common point.
143. If ∠ is a trihedral angle, int. ∠ and , , are points on edges
, , , different from , then the ray | and int. have a common
point.
144. Show that any intersection of convex sets is a convex set.

112
145. Show that the following sets are convex planes, half-planes, any open or
closed half-space and the interior of a dihedral angle.
146. Can a dihedral angle be a convex set?
147. Which of the following sets are convex:
a. a trihedral angle;
b. its interior;
c. the union of its faces;
d. the union of its interior with all its faces?
148. Let � be an open half-space bordered by plane and a closed convex
set in plane . Show that the set � is convex.
149. Show that the intersection of sphere , with a plane which passes
through , is a circle.
150. Prove that the int. , is a convex set.
151. Show that, by unifying the midpoints of the opposite edges of a
tetrahedron, we obtain concurrent lines.

113
152. Show that the lines connecting the vertices of a tetrahedron with the
centroids of the opposite sides are concurrent in the same point as the
three lines from the previous example.
153. Let be a tetrahedron. We consider the trihedral angles which have
as edges [ , [ , [ , [ , [ , [ , [ , [ , [ , [ , [ , [ . Show that
the intersection of the interiors of these 4 trihedral angles coincides with
the interior of tetrahedron [ ].
154. Show that (∀) int. [ ] ( ) | | and | | such that
‖ .
155. The interior of tetrahedron [ ] coincides with the union of segments
| | with | | and | |, and tetrahedron [ ] is equal to the
union of the closed segments [ ], when [ ] and [ ].
156. The tetrahedron is a convex set.
157. Let and convex sets. Show that by connecting segments [ ], for
which and we obtain a convex set.
158. Show that the interior of a tetrahedron coincides with the intersection of
the open half-spaces determined by the planes of the faces and the
opposite peak. Define the tetrahedron as an intersection of half-spaces.

114
Solutions
We assume that = { , } ⇒ ⊂ .
It contradicts the hypothesis = { } or = .
We assume that all the points belong to the plane for the points that
are not situated in the same plane. False!
, , , , which are not in the same plane. We assume that = { }
and are contained in the same plane and thus , , , are in the same
plane. False, it contradicts the hypothesis = lines with no
point in common.
(if all the points would , the existence of the plane and space would
be negated). Let = , (otherwise the space wouldn’t exist). Let =
, ≠ and both contain line .
We show that ≠ ′
≠ ′′
≠ .
Let ′
= { } = , ′
{
⊂
′
⊂
′
= { }
≠
{
⊂∝
, ′

115
′′ ′
= { }
≠
≠
′
′
⊂ ∝
, ′′
= ′
or
= ′′
′′
⊂ , so the lines are located on the same plane .
If
′
= { } ′
′′
= { } ′′} ′ ′′
= { }, and the three lines have a point in
common.
a. .
We assume that , , collinear such that
, ,
,
} ⇒ ⊂
– false. Therefore, the points , , are not collinear.
b. Let = .
As the planes are distinct, their intersections are:

116
=
=
=
}
If =
, , , coplanar, contrary to the hypothesis.
We suppose that , ≠ } } , , are collinear
(false, contrary to point a.). Therefore, set has a single point = { }.
We showed at the previous exercise that if , ≠ . We show
that , , are not collinear. We assume the opposite. Then,
Having three common points , and false. So , , are not collinear and
determine a plane .
, , are collinear because to the line of intersection of the two planes.

117
The planes are , ′ ; , ′ ; , ′ .
Generalization: The number of planes corresponds to the number of points on
line because ′ contains only 2 points.
Let line be given, and any point such that .
We obtain the plane = , , and let . The line ′ = , ′ ⊄ is not
thus contained in the same plane with . The desired planes are those of type
, ′, that is an infinity of planes.
a. (∀) 3 points determine a plane. Let plane ( ). We choose in this plane
| | and | | such that | |, then the line separates the points
and , but does not separate and , so it separates and ⇒ | | = ,
where | |.
Thus, the line meets 3 of the given lines. Let’s see if it can meet more.

118
We assume that
it has two points in common with the plane.
, , , coplanar – false.
Thus,
false.
We show in the same way that does not cut or , so a line meets at
most three of the given lines.
b. We consider points , , such that | |, | |, | |. These points
determine plane ( ) which obviously cuts the lines , and . does not
separate and or it does not separate or | |.

119
Let’s show that ( ) meets as well the lines , , . In the plane ( ) we
consider the triangle and the line .
As this line cuts side | |, but it does not cut | |, it must cut side | |, so
| | = { } | | ⊂ , so = { }. In the plane ( ), the line
cuts | | and does not cut | |, so cuts the side |CD|, | | = { }
⊂ = { }.
, does not separate and
separates and
} | | =
= { }.
We assume problem is solved, if } ≠ , = .
As || || = || || the bisecting line of the segment [ ].
So, to find , we proceed as follows:
1. We look for the line of intersection of planes and , d. If , the
problem hasn’t got any solution.
2. We construct the bisecting line ′ of the segment [ ] in the plane .
3. We look for the point of intersection of lines and ′. If ′, the problem
hasn’t got any solution.

120
If = = . If = , the desired intersection is , which
can be a point (the 3 planes are concurrent), the empty set (the line of intersection
of two planes is || with the third) or line (the 3 planes which pass through are
secant).
To determine , we proceed as follows:
1. We construct plane ( ) and we look for the line of intersection with , .
If (/ ), neither does .
2. We construct plane ( ) and we look for the line of intersection with , ′.
If ′ does not exist, neither does .
3. We look for the point of intersection of lines ′ and ′. The problem has
only one solution if the lines are concurrent, an infinity if they are coinciding
lines and no solution if they are parallel.
We assume the problem is solved.
a. First we assume that . , are collinear. As ′ and ′ are concurrent lines,
they determine a plane , that intersects after line ′ ′.
As
and points , , ′ are collinear ∀ .

121
b. We assume that , , are not collinear.
We notice that: ′, ′ = (plane determined by 2 concurrent lines).
= ′ and ′.
To determine we proceed as follows:
1) We determine plane ;
2) We look for the point of intersection of this plane with line , so
= { } is the desired point.
Then = ′
.
′, ′, ′ are collinear.
� and � [ ] ≠ .
Let � = | = | .
Let | | and we must show that � ∀ inside the segment.
We assume the contrary that � such that [ ] = { }
[ ] [ ] [ ] ≠ false.
, so �. The property is also maintained for the closed half-space.

122
Compared to the previous case there can appear the situation when one of the
points and or when both belong to .
If , �, | | ≠ and we show as we did above that:
If:
Let
So
Let be a plane and � , � the two half-spaces that it determines. We consider
half-space � .

123
determines on two rays, | and | where
and are in different half-spaces.
We assume
Let � be an open half-space and its margin and let = .
We choose points and – , on both sides of line
, are on one side and on the other side of and it means that only one of
them is on �.
We assume that � �. We now prove � = | .
� ⊂ |

124
Let
[ ] ≠ and are on one side and on the other side of line is
on the same side of line with |
�, so | ⊂ �.
Let σ be the considered half-space and its margin. There are more possible
cases:
In this case it is possible that:
Let
is a half-plane according to a previous problem.

126
We first assume that ≠ and | ⊂ | .
As
The hypothesis can then be written as ≠ and | ⊂ | . Let’s show that
= . By reductio ad absurdum, we assume that ≠ = and let
, so and . We draw through and a plane , such that , so
the three planes , and do not pass through this line. As has the common
point with and , it is going to intersect these planes.
which is a common point of the 3 planes. Lines and ′ determine 4 angles in
plane , having as a common peak, the interior of one of them, let int.
ℎ
̂. We consider int. ℎ
̂.
Then is on the same side with in relation to ′, so is on the same side with
in relation to | .

127
But is on the opposite side of in relation to , so is on the opposite side of
in relation to | . So | ⊄ | – false – it contradicts the hypothesis
So = .
Let be the edge of the given dihedral angle. Depending on the position of a line
in relation to a plane, there can be identified the following situations:
The ray with its origin in , so ⊂ ′ ′
̂ = ′ ′′
̂ thus an angle.
Indeed, if we assumed that ′ ′′ ≠ ′ ′′.
false – it contradicts the hypothesis.

128
Or
in this case ′ ′
̂ = ′′ - a line.
Then ′ ′
̂ = .
= , but ≠ , ≠
′ ′
̂ = thus the intersection is a line.
In this case the intersection is a closed half-plane.
is a half-plane

129
From (*) and (**),
points and are on different sides of .
is a half-plane
so they are secant planes

130
Let rays
As is interior to the dihedron formed by any half-plane passing through of the
trihedral, so
So
and in the same half-plane det. ⇒ and on the same side of (1)
and are on the same side of ′ and are on the same side of
(2).
From (1) and (2)
Let and ′′be two convex sets and ′ their intersection. Let
so the intersection is convex.
a. Let , ; ≠ | = (the line is a convex set)
⊂ , so | | ⊂ , so the plane is a convex set.

131
b. Half-planes: Let = | and , | | = . Let | | | | ⊂
| | | | = and are in the same half-plane . So
| | ⊂ and is a convex set.
Let ′ = [ . There are three situations:
1) , | – previously discussed;
2) , | | ⊂ ⊂ ′;
3) , | | ⊂ | | | ⊂ | ⊂ [ so [ is a convex set.
c. Half-spaces: Let � = | and let , � | | = .
Let | | | | ⊂ | | | | = .
Let �′ = [ . There are three situations:
1) , | previously discussed;
2) , | | ⊂ ⊂ �′;
3) , .
and so �′ is a convex set.
d. the interior of a dihedral angle:
. ′ ′ = | | and as each half-space is a convex set and their intersection
is the convex set.

132
No. The dihedral angle is not a convex set, because if we consider it as in the
previous figure ′ and ′.
Only in the case of the null or straight angle, when the dihedral angle
becomes a plane or closed half-plane, is a convex set.
a. No. The trihedral angle is not the convex set, because, if we take and
the int. ̂ determined by the int. ̂ , such that | the int. =
{ }, | |, ̂ .
So , ̂ , but | | .
b. � = , = is a convex set as an intersection of convex sets.
C) It is the same set from a. and it is not convex.
D) The respective set is [ [ [ , intersection of convex sets and, thus, it
is convex.
Let � = | and ⊂ . Let , �.
We have the following situations:

134
In triangle ABC:
In triangle DAC:
parallelogram | | and | | intersect at their midpoint .
parallelogram.
| | passes through midpoint of | |.
Thus the three lines , , are concurrent in .
Let tetrahedron and be the midpoint of | |. The centroid of the face
is on | | at a third from the base. The centroid ′ of the face is on | |
at a third from the base | |.
We separately consider ∆ . Let be the midpoint of , so is median in
this triangle and, in the previous problem, it was one of the 3 concurrent segments
in a point located in the middle of each.
Let be the midpoint of | |. We write = { ′} and = { }.

135
From (1) and (2)
′ is exactly the centroid of face , because it is situated on median | | at
a third from . We show in the same way that is exactly the centroid of face .
We’ve thus shown that and ′ pass through point from the previous
problem.
We choose faces and and mark by ′′ the centroid of face , we
show in the same way that and ′′ pass through the middle of the segment
| | | | ≡ | |, | | ≡ | | thus also through point , etc.
We mark planes = , = , = , = .
Let be the intersection of the interiors of the 4 trihedral angles.
We show that:
= int. [ ], by double inclusion.
1. int. ̂ int. ̂ int. ̂ int. ̂ | | and
| | | | and | and | and |
| | int. [ ]. So [ ].
2. Following the inverse reasoning we show that [ ] ⊂ from where the
equality.

137
Let
such that [ ]. But [ ] ⊂ [ ] because [ ] [ ] and the
surface of the triangle is convex. So [ ] such that
and the tetrahedron is a convex set.
Note: The tetrahedron can be regarded as the intersection of four closed half-
spaces which are convex sets.
Let ℳ be the union of the segments [ ] with ℳ and ℳ .
Let , ’ ℳ ∀ ℳ and ℳ such that [ ];
′ ℳ and ′ ℳ such that ′ [ ′ ′].
From , ′ ℳ [ ′]′
ℳ which is a convex set.
From , ′ ℳ [ ′] ℳ which is a convex set.
The union of all the segments [ ] with [ ′] and [ ′] is tetrahedron
[ ′ ′] ⊂ ℳ.
So from , ′ ℳ | ’| ⊂ ℳ, so set ℳ is convex.
The interior of the tetrahedron coincides with the union of segments | |,
| | and | |, that is int. [ ] = {| | | |, | |}.
Let’s show that:

138
1. Let
2. Let
If we assume | | | | ≠ and are in different half-
spaces in relation to , , false (it contradicts the hypothesis).
So
and the second inclusion is proved.
As regarding the tetrahedron: [ ] = {[ ] [ ] and [ ]}.

139
If
= , [ ], [ ] describes face [ ]
= , [ ], [ ] describes face [ ]
= , [ ], [ ] describes face [ ].
Because the triangular surfaces are convex sets and along with their two points ,
, segment [ ] is included in the respective surface.
So, if we add these two situations to the equality from the previous case, we
obtain:

140
Lines and Planes
159. Let , ′
be two parallel lines. If the line is parallel to a plane , show
that ′
|| or ′
⊂ .
160. Consider a line , parallel to the planes and , which intersects after the
line . Show that ‖ .
161. Through a given line , draw a parallel plane with another given line ′
.
Discuss the number of solutions.
162. Determine the union of the lines intersecting a given line and parallel
to another given line ′ ′
.
163. Construct a line that meets two given lines and that is parallel to a third
given line. Discuss.
164. If a plane intersects the secant planes after parallel lines, then is
parallel to line .
165. A variable plane cuts two parallel lines in points and . Find the
geometrical locus of the middle of segment [ ].

141
166. Two lines are given. Through a given point, draw a parallel plane with
both lines. Discuss.
167. Construct a line passing through a given point, which is parallel to a
given plane and intersects a given line. Discuss.
168. Show that if triangles and ′ ′ ′, located in different planes, have
′ ′, ′ ′ and ′ ′, then lines ′, ′, ′ are concurrent or
parallel.
169. Show that, if two planes are parallel, then a plane intersecting one of
them after a line cuts the other one too.
170. Through the parallel lines and ′
we draw the planes and ′
distinct
from ( , ′
). Show that ′
or ′
.
171. Given a plane , a point and a line ⊂ .
a. Construct a line ′
such that ′
⊂ , ′
and ′
.
b. Construct a line through included in , which forms with an angle
of a given measure . How many solutions are there?
172. Show that relation defined on the set of planes is an equivalence
relation. Define the equivalence classes.

142
173. Consider on the set of all lines and planes the relation or = ,
where and are lines or planes. Have we defined an equivalence
relation?
174. Show that two parallel segments between parallel planes are concurrent.
175. Show that through two lines that are not contained in the same plane, we
can draw parallel planes in a unique way. Study also the situation when the
two lines are coplanar.
176. Let and be two parallel planes, , , and is a parallel line with
and . Lines , , , cut plane respectively in , , , . Show
that these points are the vertices of a parallelogram.
177. Find the locus of the midpoints of the segments that have their
extremities in two parallel planes.

143
Solutions
Let ′′
} ′′
′
′′ } ′ ′′
}
′′
or ′
⊂ .
Let ⇒ . We draw through A, ′
.
, ′ } ′
⊂
, ′ } ′
⊂
}
′
⊂ ⊂
=
}
′
=
′ }
a. If ′ there is only one solution and it can be obtained as it follows:
Let . In the plane ( , ′′
) we draw ′′ ′
. The concurrent lines and ′′
determine plane . As ′′
, in the case of the non-coplanar lines.

144
b. If ′ || d'', ( ) infinite solutions. Any plane passing through is parallel to
′′
, with the exception of plane ( , ′).
c. ′, but they are coplanar ( ) solutions.
Let , we draw through , ′. We write = , . As ′ ′ .
Let , arbitrary .
We draw
′
,
′
} ⊂ , so all the parallel lines to ′ intersecting are
contained in plane .
Let ⊂ , ′
= , so (∀) parallel to ′
from intersects . Thus, the
plane represents the required union.

145
We draw through such that
′
′
} . According to previous
problem: = { }. Therefore,
a. If , the plane is unique, and if ≠ , the solution is unique.
b. If , ( )
≠
, because it would mean that we can draw through a
point two parallel lines , to the same line . So there is no solution.
c. If and ≠ , all the parallel lines to cutting are on the plane
and none of them can intersect , so the problem has no solution.
d. If ⊂ , ≠ , let = { }, and the required line is parallel to
drawn through one solution.
e. If ⊂ , . The problem has infinite solutions, (∀) || to which cuts ,
also cuts .

146
The problem is reduced to the geometrical locus of the midpoints of the
segments that have extremities on two parallel lines. is such a point | | = | |.
We draw ⊥ ⇒ ?̂ = ̂ � = � | | ≡ | | || ||?
the geometrical locus is the parallel to and drawn on the mid-distance
between them. It can also be proved vice-versa.

147
Let . In plane , we draw ′
, ′
. In plane ( ) we draw
′
, ′
. We note = ′ ′
the plane determined by two concurrent lines.
the only solution.
Let , , .
= ′
= ′
In this case ′
= ′
= and infinite planes pass through ;
} , are parallel lines with (∀) of the planes passing through .
The problem has infinite solutions. But or , the problem has no
solution because the plane can’t pass through a point of a line and be parallel to
that line.
Let be the given point, the given plane and the given line.
a. We assume that , = { }. Let plane which has a common point
with ⇒ = ′.
We draw in plane through point a parallel line to ′.

148
is the required line.
b. , ≠ .
Let dA = ’} ⇒ ′
All the lines passing through and intersecting are contained in plane . But
all these lines also cut ′ , so they can’t be parallel to . There is no solution.
c. , = .
Let and line ⊂ ; = = , ∀ .
The problem has infinite solutions.
and ′ ′ ′ are distinct planes, thus the six points , , , ′, ′, ′ can’t be
coplanar.
′ ’ , , ′, ′ are coplanar.

149
The points are coplanar four by four, that is ′ , ′ ′ , ′ ′ , and
determine four distinct planes. If we assumed that the planes coincide two by two,
it would result other 6 coplanar points and this is false.
In plane ′ ′, lines ′, ′ can be parallel or concurrent.
First we assume that:
is a common point to the 3 distinct planes, but the intersection of 3 distinct
planes can be only a point, a line or . It can’t be a line because lines
are distinct if we assumed that two of them coincide, the 6 points would be
coplanar, thus there is no common line to all the three planes. There is one
possibility left, that is they have a common point and from
We assume = plane drawn through ⊂ , false.
So = { }.
Hypothesis: , = .
Conclusion: = .
We assume that = .
Let

150
because from a point we can draw only one parallel plane with the given plane.
But this result is false, it contradicts the hypothesis = so = .
Hypothesis: ′; ⊂ ; ′ ⊂ ′; , ′ ≠ ′
.
Conclusion: ′ or ′′ .
As , ′ ≠ ′
≠ ′.
If ′
= ′.
If ′
= ′
= ′′
.
If
′′
′ ⊂ ′
′
⊂
} ′′‖ .
a. If , then ′ = . If , we draw through , ′ .

255 Compiled And Solved Problems In Geometry And Trigonometry

255 Compiled And Solved Problems In Geometry And Trigonometry

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255 Compiled And Solved Problems In Geometry And Trigonometry