PROBLEMS IN PLANE AND SOLID
GEOMETRY
v.1 Plane Geometry
Viktor Prasolov
translated and edited by Dimitry Leites
Abstract. This book has no equal. The priceless treasures of elementary geometry are
nowhere else exp osed in so complete and at the same time transparent form. The short
solutions take barely 1.5 − 2 times more space than the formulations, while still remaining
complete, with no gaps whatsoever, although many of the problems are quite difficult. Only
this enabled the author to squeeze about 2000 problems on plane geometry in the book of
volume of ca 600 pages thus embracing practically all the known problems and theorems of
elementary geometry.
The book contains non-standard geometric problems of a level higher than that of the
problems usually offered at high school. The collection consists of two parts. It is based on
three Russian editions of Prasolov’s books on plane geometry.
The text is considerably modified for the English edition. Many new problems are added
and detailed structuring in accordance with the methods of solution is adopted.
The book is addressed to high school students, teachers of mathematics, mathematical
clubs, and college students.
Contents
Editor’s preface 11
From the Author’s preface 12
Chapter 1. SIMILAR TRIANGLES 15
Background 15
Introductory problems 15
§1. Line segments intercepted by parallel lines 15
§2. The ratio of sides of similar triangles 17
§3. The ratio of the areas of similar triangles 18
§4. Auxiliary equal triangles 18
* * * 19
§5. The triangle determined by the bases of the heights 19

∗ ∗∗ 61
§6. Application of the theorem on triangle’s heights 61
§7. Areas of curvilinear figures 62
§8. Circles inscribed in a disc segment 62
§9. Miscellaneous problems 63
§10. The radical axis 63
Problems for independent study 65
Solutions 65
CHAPTER 4. AREA 79
Background 79
Introductory problems 79
§1. A median divides the triangle
into triangles of equal areas 79
§2. Calculation of areas 80
§3. The areas of the triangles into which
a quadrilateral is divided 81
§4. The areas of the parts into which
a quadrilateral is divided 81
§5. Miscellaneous problems 82
* * * 82
§6. Lines and curves that divide figures
into parts of equal area 83
§7. Formulas for the area of a quadrilateral 83
§8. An auxiliary area 84
§9. Regrouping areas 85
Problems for independent study 86
Solutions 86
CHAPTER 5. TRIANGLES 99
Background 99
Introductory problems 99

* * * 138
§2. Quadrilaterals 139
§3. Ptolemy’s theorem 140
§4. Pentagons 141
§5. Hexagons 141
§6. Regular polygons 142
* * * 142
* * * 143
§7. The inscribed and circumscribed p olygons 144
* * * 144
§8. Arbitrary convex polygons 144
§9. Pascal’s theorem 145
Problems for independent study 145
Solutions 146
Chapter 7. LOCI 169
Background 169
Introductory problems 169
§1. The locus is a line or a segment of a line 169
* * * 170
§2. The locus is a circle or an arc of a circle 170
* * * 170
§3. The inscribed angle 171
§4. Auxiliary equal triangles 171
§5. The homothety 171
§6. A method of loci 171
§7. The locus with a nonzero area 172
§8. Carnot’s theorem 172
§9. Fermat-Apollonius’s circle 173
Problems for independent study 173
Solutions 174

§8. Broken lines inside a square 209
§9. The quadrilateral 210
§10. Polygons 210
* * * 211
§11. Miscellaneous problems 211
* * * 211
Problems for independent study 212
Supplement. Certain inequalities 212
Solutions 213
Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 235
§1. Medians 235
§2. Heights 235
§3. The bisectors 235
§4. The lengths of sides 236
§5. The radii of the circumscribed, inscribed and escribed circles 236
§6. Symmetric inequalities between the angles of a triangle 236
§7. Inequalities between the angles of a triangle 237
§8. Inequalities for the area of a triangle 237
* * * 238
§9. The greater angle subtends the longer side 238
§10. Any segment inside a triangle is shorter than the longest side 238
§11. Inequalities for right triangles 238
§12. Inequalities for acute triangles 239
§13. Inequalities in triangles 239
Problems for independent study 240
Solutions 240
Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM 255
CONTENTS 7
Background 255
Introductory problems 255

§2. Inner product. Relations 290
§3. Inequalities 291
§4. Sums of vectors 292
§5. Auxiliary projections 292
§6. The method of averaging 293
§7. Pseudoinner product 293
Problems for independent study 294
Solutions 295
Chapter 14. THE CENTER OF MASS 307
Background 307
§1. Main properties of the center of mass 307
§2. A theorem on mass regroupping 308
§3. The moment of inertia 309
§4. Miscellaneous problems 310
§5. The barycentric coordinates 310
8 CONTENTS
Solutions 311
Chapter 15. PARALLEL TRANSLATIONS 319
Background 319
Introductory problems 319
§1. Solving problems with the aid of parallel translations 319
§2. Problems on construction and loci 320
* * * 320
Problems for independent study 320
Solutions 320
Chapter 16. CENTRAL SYMMETRY 327
Background 327
Introductory problems 327
§1. Solving problems with the help of a symmetry 327
§2. Properties of the symmetry 328

Chapter 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 359
Background 359
Introductory problems 359
§1. Homothetic polygons 359
§2. Homothetic circles 360
§3. Costructions and loci 360
CONTENTS 9
* * * 361
§4. Composition of homotheties 361
§5. Rotational homothety 361
* * * 362
* * * 362
§6. The center of a rotational homothety 362
§7. The similarity circle of three figures 363
Problems for independent study 364
Solutions 364
Chapter 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT 375
Background 375
§1. The least and the greatest angles 375
§2. The least and the greatest distances 376
§3. The least and the greatest areas 376
§4. The greatest triangle 376
§5. The convex hull and the base lines 376
§6. Miscellaneous problems 378
Solutions 378
Chapter 21. DIRICHLET’S PRINCIPLE 385
Background 385
§1. The case when there are finitely many points, lines, etc. 385
§2. Angles and lengths 386
§3. Area 387

§1. Systems of points 437
§2. Systems of segments, lines and circles 437
§3. Examples and counterexamples 438
Solutions 438
Chapter 27. INDUCTION AND COMBINATORICS 445
§1. Induction 445
§2. Combinatorics 445
Solutions 445
Chapter 28. INVERSION 449
Background 449
§1. Properties of inversions 449
§2. Construction of circles 450
§3. Constructions with the help of a compass only 450
§4. Let us perform an inversion 451
§5. Points that lie on one circle and circles passing through one point 452
§6. Chains of circles 454
Solutions 455
Chapter 29. AFFINE TRANSFORMATIONS 465
§1. Affine transformations 465
§2. How to solve problems with the help of affine transformations 466
Solutions 466
Chapter 30. PROJECTIVE TRANSFORMATIONS 473
§1. Projective transformations of the line 473
§2. Projective transformations of the plane 474
§3. Let us transform the given line into the infinite one 477
§4. Application of projective maps that preserve a circle 478
§5. Application of projective transformations of the line 479
§6. Application of projective transformations of the line in problems on construction 479
§7. Impossibility of construction with the help of a ruler only 480
Solutions 480

Part 2 includes more recent topics, geometric t ransformations and problems more suitable
for contests and for use in mathematical clubs. The problems cover cuttings, colorings, the
pigeonhole (or Dirichlet’s) principle, induction, and so on.
Part 3 is devoted to solid geometry.
A rather detailed table of contents serves as a guide in the sea of geometric problems. It
helps the experts to easily find what they need while the uninitiated can quickly learn what
exactly is that they are interested in in geometry. Splitting the book into small sections (5
to 10 problems in each) made the book of interest to the readers of various levels.
FOR THE ENGLISH VERSION of the book about 150 new problems are already added
and several hundred more of elementary and intermideate level problems will be added to
make the number of more elementary problems sufficient to use the book in the ordinary
school: the Russian editions are best suited for coaching for a mathematical Olympiad than
for a regular class work: the level of difficulty increases rather fast.
Problems in each section are ordered difficulty-wise. The first problems of the sections
are simple; they are a match for many. Here are some examples:
1
Here are a few samples: v. 96, n. 5, 1989, p. 429–431 (here the main idea of the solution is the
right illustration — precisely the picture from the back cover of the 1st Russian edition of Problems in Solid
Geometry, Fig. to Problem 13.22); v. 96, n. 6, p. 527, Probl. E3192 corresponds to Problems 5.31 and
18.20 of Problems in Plane Geometry — with their two absolutely different solutions, the one to Problem
5.31, unknown to AMM, is even more interesting.
12 CONTENTS
Plane 1.1. The bases of a trapezoid are a and b. Find the length of the segment that
the diagonals of the trapezoid inter sept on the trapezoid’s midline.
Plane 1.52. Let AA
1
and BB
1
be the altitudes of △ABC. Prove that △A
1

The final problems of the sections are usually borrowed from scientific journals. Here are
some examples:
Plane 10.20. Prove that l
a
+ l
b
+ m
c
≤
√
3p, where l
a
, l
b
are the lengths of the bisectors
of the angles ∠A and ∠B of the triangle △ABC, m
c
is the length of the median of the side
AB, and p is the semiperimeter.
Plane 19.55. Let O be the center of the circle inscribed in △ABC, K the Lemoine’s
point, P and Q Brocard’s points. Prove that P and Q belong to the circle with diameter
KO and that OP = OQ.
Plane 22.29. The numbers α
1
, . . . , α
n
, whose sum is equal to (n−2)π, satisfy inequalities
0 < α
i
< 2π. Prove that there exists an n- gon A

but also as a self-guide for those who wish (or have no other choice but) to study geometry
FROM THE AUTHOR’S PREFACE 13
independently. Detailed headings are provided for the reader’s convenience. Problems in the
two parts of Plane are spread over 29 Chapters, each Chapter comprising 6 to 14 sections.
The classification is based on the methods used to solve geometric problems. The purpose of
the division is basically to help the reader find his/her bearings in t his large array of problems.
Otherwise the huge number of problems might be somewhat depressingly overwhelming.
Advice and comments given by Academician A. V. Pogorelov, and Professors A. M. Abramov,
A. Yu. Vaintrob, N. B. Vasiliev, N. P. Dolbilin, and S . Yu. Orevkov were a great help to me
in preparing the first Soviet edition. I wish to express my sincere gratitude to all of them.
To save space, sections with background only contain the material directly pertinent to
the respective chapter. It is collected just to remind the reader of notations. Therefore, the
basic elements of a triangle are only defined in chapter 5, while in chapter 1 we assume that
their definition is known. For the reader’s convenience, cross references in this translation
are facilitated by a very detailed index.

Chapter 1. SIMILAR TRIANGLES
Background
1) Triangle ABC is said to be similar to triangle A
1
B
1
C
1
(we write △ABC ∼ △A
1
B
1
C
1

c) ∠ABC = ∠A
1
B
1
C
1
and ∠BAC = ∠B
1
A
1
C
1
.
2) Triangles AB
1
C
1
and AB
2
C
2
cut off from an angle with vertex A by parallel lines are
similar and AB
1
: AB
2
= AC
1
: AC
2

and B
1
B
2
. . . B
n
are called similar if A
1
A
2
: A
2
A
3
: ··· : A
n
A
1
=
B
1
B
2
: B
2
B
3
: ··· : B
n
B

= AH · BH.
3. Prove that the medians of a triangle meet at one point and this point divides each
median in the ratio of 2 : 1 counting from the vertex.
4. On side BC of △ABC point A
1
is taken so that BA
1
: A
1
C = 2 : 1. What is the
ratio in which median CC
1
divides segment AA
1
?
5. Square PQRS is inscribed into △ABC so that vertices P and Q lie on sides AB and
AC and vertices R and S lie on BC. Express the length of the square’s side through a and
h
a
.
§1. Line segments intercepted by parallel lines
1.1. Let the lengths of bases AD and BC of trapezoid ABCD be a and b (a > b).
15
16 CHAPTER 1. SIMILAR TRIANGLES
a) Find the length of the segment that the diagonals intercept on the midline.
b) Find the length of segment MN whose endpoints divide AB and CD in the ratio of
AM : MB = DN : NC = p : q.
1.2. Prove that the midpoints of the sides of an arbitrary quadrilateral are vertices of
a parallelogram. For what quadrilaterals this parallelogram is a rectangle, a r hombus, a
square?

B
1
 AB.
1.5. The straight line which connects the intersection point P of the diagonals in quadri-
lateral ABCD with the intersection point Q of the lines AB and CD bisects side AD. Prove
that it also bisects BC.
1.6. A point P is taken on side AD of parallelogram ABCD so that AP : AD = 1 : n;
let Q be the intersection point of AC and BP . Prove that AQ : AC = 1 : (n + 1).
1.7. The vertices of parallelogram A
1
B
1
C
1
D
1
lie on the sides of parallelogram ABCD
(point A
1
lies on AB, B
1
on BC, etc.). Prove that the centers of the two parallelograms
coincide.
1.8. Point K lies on diagonal BD of parallelogram ABCD. Straight line AK intersects
lines BC and CD at points L and M, respectively. Prove that AK
2
= LK · KM .
1.9. One of th e diagonals of a quadrilateral inscribed in a circle is a diameter of the
circle. Prove that (the lengths of) the projections of the opposite sides of the quadrilateral
on the other diagonal are equal.

1
.
b) Points A, B, and C lie on one straight line and A
1
, B
1
, and C
1
are such that AB
1

BA
1
, AC
1
 CA
1
, and BC
1
 CB
1
. Prove that A
1
, B
1
and C
1
lie on one line.
1.13. In △ABC bisectors AA
1

side is equal to b. Calculate the radius of the circumscribed circle.
1.19. A straight line passing through vertex A of square ABCD intersects side CD at
E and line BC at F . Prove that
1
AE
2
+
1
AF
2
=
1
AB
2
.
1.20. Given points B
2
and C
2
on h eights BB
1
and CC
1
of △ABC such that AB
2
C =
AC
2
B = 90
◦

2
+ bc = c
2
.
1.26. The endpoints of segments AB and CD are gliding along the sides of a given angle,
so that straight lines AB and CD are moving parallelly (i.e., each line moves parallelly to
itself) and segments AB and CD intersect at a point, M. Prove that the value of
AM·BM
CM·D M
is
a constant.
1.27. Through an arbitrary point P on side AC of △ABC straight lines are drawn
parallelly to the triangle’s medians AK and CL. The lines intersect BC and AB at E and
F , respectively. Prove that AK and CL divide EF into three equal parts.
1.28. Point P lies on the bisector of an angle with vertex C. A line passing through P
intercepts segments of lengths a and b on the angle’s legs. Prove that the value of
1
a
+
1
b
does
not depend on the choice of the line.
1.29. A semicircle is constructed outwards on side BC of an equilateral triangle ABC
as on the diameter. Given points K and L that divide the semicircle into three equal arcs,
prove that lines AK and AL divide BC into three equal parts.
1.30. Point O is the center of the circle inscribed in △ABC. On sides AC and BC
points M and K, respectively, are selected so that BK ·AB = BO
2
and AM · AB = AO

allelly to the sides of the triangle. The lines divide the triangle into six parts, three of
which are triangles of areas S
1
, S
2
and S
3
. Prove that the area of △ABC is equal to

√
S
1
+
√
S
2
+
√
S
3

2
.
1.36. Prove that the area of a triangle whose sides are equal to the medians of a triangle
of area S is equal to
3
4
S.
1.37. a) Prove that the area of t he quadrilateral formed by the midpoints of the sides of
convex quadrilateral ABCD is half that of ABCD.

and D
1
D
2
are equal and perpendicular to each other.
1.42. Consider an isosceles right triangle ABC with CD = CE and points D and E on
sides CA and CB, respectively. Extensions of perpendiculars dropped from D and C to AE
intersect the hypotenuse AB at K and L. Prove that KL = LB.
1.43. Consider an inscribed quadrilateral ABCD. The lengths of sides AB, BC, CD,
and DA are a, b, c, and d, respectively. Rectangles are constructed outwards on the sides of
§5. THE TRIANGLE DETERMINED BY THE BASES OF THE HEIGHTS 19
the quadrilateral; the sizes of the rectangles are a × c, b × d, c × a and d × b, respectively.
Prove that the centers of the rectangles are vertices of a rectangle.
1.44. Hexagon ABCDEF is inscribed in a circle of radius R centered at O; let AB =
CD = EF = R. Prove that the intersection points, other than O, of the pairs of circles
circumscribed about △BOC, △DOE and △F OA are the vertices of an equilateral triangle
with side R.
* * *
1.45. Equilateral triangles BCK and DCL are constructed outwards on sides BC and
CD of parallelogram ABCD. Prove that AKL is an equilateral triangle.
1.46. Squares are constructed outwards on the sides of a parallelogram. Prove that their
centers form a square.
1.47. Isosceles triangles with angles 2α, 2β and 2γ at vertices A
′
, B
′
and C
′
are con-
structed outwards on the sides of triangle ABC; let α + β + γ = 180

1
= ∠B
1
= 90
◦
, ∠ABC
1
= ∠ACB
1
= ϕ; let M be the
midpoint of BC. Prove that MB
1
= MC
1
and ∠B
1
MC
1
= 2ϕ.
b) Equilateral triangles are constructed outwards on the sides of △ABC. Prove that the
centers of the triangles constructed form an equilateral triangle whose center coincides with
the intersection point of the medians of △ABC.
1.50. Isosceles triangles AC
1
B and AB
1
C with an angle ϕ at the vertex are constructed
outwards on the unequal sides AB and AC of a scalene triangle △ABC.
a) L et M be a point on median AA
1

B
1
C ∼ △ABC. What
is the similarity coefficient?
1.53. Height CH is dropped from vertex C of acute triangle ABC and perpendiculars
HM and HN are dropped to sides BC and AC, respectively. Prove that △MNC ∼ △ABC.
1.54. In △ABC heights BB
1
and CC
1
are drawn.
a) Prove that the tangent at A to the circumscribed circle is parallel to B
1
C
1
.
b) Prove that B
1
C
1
⊥ OA, where O is the center of the circumscribed circle.
1.55. Points A
1
, B
1
and C
1
are taken on the sides of an acute triangle ABC so that
segments AA
1

1
and B
1
are taken on sides AB, BC and CA, respectively, of acute triangle
ABC. Prove that if ∠B
1
A
1
C = ∠BA
1
C
1
, ∠A
1
B
1
C = ∠AB
1
C
1
and ∠A
1
C
1
B = ∠AC
1
B
1
,
then points A

1
 AB and B
1
C
1
 BC, then A
1
C
1
 AC.
1.59. Let p be the semiperimeter of acute triangle ABC and q the semiperimeter of the
triangle formed by the bases of the heights of △ABC. Prove that p : q = R : r, where R
and r are the radii of the circumscribed and the inscribed circles, respectively, of △ABC.
§6. Similar figures
1.60. A circle of radius r is inscribed in a triangle. The straight lines tangent to the
circle and parallel to the sides of the triangle are drawn; the lines cut three small triangles
off the triangle. Let r
1
, r
2
and r
3
be the radii of the circles inscribed in the small triangles.
Prove that r
1
+ r
2
+ r
3
= r.

b) Prove that BF DE is a rectangle.
1.65. Perpendiculars MQ and MP are dropped from an arbitrary point M of the circle
circumscribed about rectangle ABCD to the rectangle’s two opp osite sides; the perpendic-
ulars MR and MT are dropped to the extensions of the other two sides. Prove that lines
P R ⊥ QT and the intersection point of PR and QT belongs to a diagonal of ABCD.
1.66. Two circles enclose non-intersecting areas. Common tangent lines to the two
circles, one external and one internal, are drawn. Consider two straight lines each of which
passes through the tangent points on one of the circles. Prove that the intersection point of
the lines lies on the straight line that connects the centers of the circles.
Problems for independent study
1.67. The (length of the) base of an isosceles triangle is a quarter of its perimeter. From
an arbitrary point on the base straight lines are drawn parallel to the sides of the triangle.
How many times is the perimeter of the triangle greater than that of the parallelogram?
1.68. The diagonals of a trapezoid are mutually perpendicular. The intersection point
divides the diagonals into segments. Prove that the product of the lengths of the trapezoid’s
bases is equal to the sum of the products of the lengths of the segments of one diagonal and
those of another diagonal.
1.69. A straight line is drawn through the center of a unit square. Calculate the sum of
the squared distances between the four vertices of the square and the line.
SOLUTIONS 21
1.70. Points A
1
, B
1
and C
1
are symmetric to the center of the circumscribed circle of
△ABC through the triangle’s sides. Prove that △ABC = △A
1
B

point of the diagonals. The endpoints of EF lie on AB and CD. Prove that AE : CF =
AO : CO.
1.77. Three straight lines parallel to the sides of the given triangle cut three triangles off
it leaving an equilateral hexagon. Find the length of th e side of the hexagon if the lengths
of the triangle’s sides are a, b and c.
1.78. Three straight lines parallel to the sides of a triangle meet at one point, the sides
of the triangle cutting off the line segments of length x each. Find x if the lengths of the
triangle’s sides are a, b and c.
1.79. Point P lies inside △ABC and ∠ABP = ∠ACP . On straight lines AB and AC,
points C
1
and B
1
are taken so that BC
1
: CB
1
= CP : BP . Prove that one of the diagonals
of the parallelogram whose two sides lie on lines BP and CP and two other sides ( or their
extensions) pass through B
1
and C
1
is parallel to BC.
Solutions
1.1. a) Let P and Q be the midpoints of AB and CD; let K and L be the intersection
points of PQ with the diagonals AC and BD, respectively. Then P L =
a
2
and PK =

1
by O. In △B
1
BC draw segment
A
1
A
2
so that A
1
A
2
 BB
1
. Then
B
1
C
B
1
A
2
= 1 + p and so AO : OA
1
= AB
1
: B
1
A
2

: A
1
B = CP : 2P C
1
. Similarly, CB
1
: B
1
A = CP : 2P C
1
= CA
1
: A
1
B.
1.5. Point P lies on the median QM of △AQD (or on its extension). It is easy to
verify that the solution of Problem 1.4 remains correct also for the case when P lies on the
extension of the median. Consequently, BC  AD.
1.6. We have AQ : QC = AP : BC = 1 : n because △AQP ∼ △CQB. So AC =
AQ + QC = (n + 1)AQ.
1.7. The center of A
1
B
1
C
1
D
1
being the midpoint of B
1

and, since P is the midpoint of BD,
it follows that BA
1
= DC
1
.
1.10. We see that BO : OD = DP : P B = k, because BO =P D. Let BC = 1. Then
AD = k and ED =
1
k
. So k = AD = AE + ED = 1 +
1
k
, th at is k
2
= 1 + k. Finally, observe
that k
2
= AD
2
and 1 + k = BC
2
+ BC · AD.
1.11. Let C, D, E and F be the midp oints of sides AO, OB, BM and MA, respectively,
of quadrilateral AOMB. Since AB = MO = R, where R is the radius of the given circle,
CDEF is a rhombus by Problem 1.2. Hence, CE ⊥ DF .
1.12. a) If the lines containing the given points are parallel, then the assertion of the
problem is obviously true. We assume that the lines meet at O. Then OA : OB = OB
1
: OA

: C
1
A. Since △CB
1
D ∼ △EB
1
A, points A
1
, B
1
and C
1
lie on the same line.
1.13. A point that lies on the bisector of an angle is equidistant from the angle’s legs.
Let a be the distance from point A
1
to lines AC and AB, let b be the distance from point B
1
to lines AB and BC. Further, let A
1
M : B
1
M = p : q, where p + q = 1. Then the distances
SOLUTIONS 23
from point M to lines AC and BC are equal to qa and pb, respectively. On the other hand,
by Problem 1.1 b) the distance from point M to line AB is equal to qa + pb.
1.14. Let the line that passes through the center O of the given rectangle parallel to BC
intersect line segment QN at point K (Fig. 3).
Figure 3 (Sol. 1.14)
Since MO  P C, it follows that QM : MP = QO : OC and, since KO  BC, it follows

QS
1
meet at N; let N
1
and N
2
be the intersection points of the line that passes through N
parallel to DD
1
with segments P R and QS, resp ect ively.
Then
−−→
N
1
N = β
−−→
RR
1
= αβ
−−→
DD
1
and
−−→
N
2
N = α
−−→
SS
1

1
C = BC we get
BA
1
=
ac
b+c
. Since BO is the bisector of triangle ABA
1
, it follows that AO : OA
1
= AB :
BA
1
= (b + c) : a.
1.18. Let O be the center of the circumscribed circle of isosceles triangle ABC, let B
1
be the midpoint of base AC and A
1
the midpoint of the lateral side BC. Since △BOA
1
∼
△BCB
1
, it follows that BO : BA
1
= BC : BB
1
and, therefore, R = BO =
a

ϕ
AB
2
=
1
AB
2
.
1.20. It is easy to verify that AB
2
2
= AB
1
· AC = AC
1
· AB = AC
2
2
.
1.21. a) Since BQ : QM = BN : AM = BK : AK, we have: KQ  AM.
b) Let O be the center of the inscribed circle. Since ∠CBA + ∠BAD = 180
◦
, it follows
that ∠ABO + ∠BAO = 90
◦
. Therefore, △AKO ∼ △OKB, i.e., AK : KO = OK : KB.
Consequently, AK·KB = KO
2
= R
2

Therefore,
AB
AE
+
AD
AF
=
AB
′
AG
+
AD
′
AG
=
CD
′
+ AD
′
AG
=
AC
AG
.
1.24. Let us drop from vertex B perpendicular BG to AC (Fig. 4).
Figure 4 (Sol. 1.24)
Since triangles ABG and ACE are similar, AC · AG = AE ·AB. Lines AF and CB are
parallel, consequently, ∠GCB = ∠CAF . We also infer that right triangles CBG and ACF
are similar and, therefore, AC ·CG = AF · BC. Summing the equalities obtained we get
AC · (AG + CG) = AE ·AB + AF ·BC.

remains a constant.
1.27. Let medians meet at O; denote the intersection points of median AK with lines
F P and FE by Q and M, respectively; denote the intersection points of median CL with
lines EP and F E by R and N, respectively (Fig. 5).
Clearly, F M : F E = F Q : F P = LO : LC = 1 : 3, i.e., F M =
1
3
F E. Similarly,
EN =
1
3
F E.
SOLUTIONS 25
Figure 5 (Sol. 1.27)
1.28. Let A and B be th e intersection points of the given line with the angle’s legs.
On segments AC and BC, take points K and L, respectively, so that P K  BC and
P L  AC. Since △AKP ∼ △P LB, it follows that AK : KP = PL : LB and, therefore,
(a −p)(b −p) = p
2
, where p = P K = PL. Hence,
1
a
+
1
b
=
1
p
.
1.29. Denote the midpoint of side BC by O and the intersection points of AK and AL

◦
, angles ∠AEB and ∠BEC cannot be different angles
of similar triangles ABE and BEC, i.e., the angles are equal and BE is a perpendicular.
Two cases are possible: either ∠ABE = ∠CBE or ∠ABE = ∠BCE. The first case
should be discarded because in this case △ABE = △CBE.
In the second case we have ∠ABC = ∠ABE + ∠CBE = ∠ABE + ∠BAE = 90
◦
. In
right triangle ABC the ratio of the legs’ lengths is equal to 1 :
√
3; hence, the angles of
triangle ABC are equal to 90
◦
, 60
◦
, 30
◦
.
1.33. We have
S
BDEF
2S
ADE
=
S
BDE
S
ADE
=
DB

2
: b
2
. Since
S
EM N
− S
EBC
= S
MBCN
= S
MADN
= S
EAD
− S
EM N
, it follows that x
2
− a
2
= b
2
− x
2
, i.e.,
x
2
=
1
2