Sharygin Olympiads, the best

The Sharygin Olympiads are a prestigious series of geometry competitions named after Igor Fedorovich Sharygin (13/02/1937 – 12/03/2004), a renowned Soviet and Russian mathematician, teacher, and popularizer of science. Sharygin specialized in elementary geometry, authored numerous textbooks, and created many elegant problems. He also led the mathematics section of the Russian Soros Olympiads.

In his memory, Russia annually hosts the Geometry Olympiad for high school students, known as the Sharygin Olympiad. It consists of two stages: a correspondence round held from January to March, and a final round. The competition is open to Russian students in grades 8 to 11 and is also accessible to international high-school students.

Usually final round hosted near Dubna in Moscow region in format similar to IMO format.

In 2022 Cheenta Academy has hosted the final round of the Sharygin Geometrical Olympiad in Kolkata, India, in collaboration with the Moscow Center for Lifelong Mathematical Education. During the final round, participants present their solutions orally to a jury, using drafts and diagrams to support their reasoning.

Selected problems from these Olympiads are published, labeled by the year and serial number. The published solutions may differ from the original ones proposed during the contest.

1 2025 I Problem 1
2 2025 I Problem 2
3 2025 I Problem 3
4 2025 I Problem 4
5 2025 I Problem 5
6 2025 I Problem 6
7 2025 I Problem 7
8 2025 I Problem 8
9 2025 I Problem 9
10 2025 I Problem 10
11 2025 I Problem 11
12 2025 I Problem 12
13 2025 I Problem 13
14 2025 I Problem 14
15 2025 I Problem 15
16 2025 I Problem 16
17 2025 I Problem 17
18 2025 I Problem 18
19 2025 I Problem 19
20 2025 I Problem 20
21 2025 I Problem 21
22 2025 I Problem 22
23 2025 I Problem 23
24 2025 I Problem 24
25 Proof
26 2024 II Problem 6
27 2024 II Problem 2(8)
28 2024 II Problem 4(8)
29 2024 II Problem 3(9)
30 2024 II Problem 4(9)
31 2024 II Problem 5(9)
32 2024 II Problem 7(9)
33 2024 II Problem 7(10)
34 2024 I Problem 2
35 2024 I Problem 8
36 2024 I Problem 9
37 2024 I Problem 12
38 2024 I Problem 14
39 2024 I Problem 15
40 2024 I Problem 16
41 2024 I Problem 17
42 2024 I Problem 18
43 2024 I Problem 19
44 2024 I Problem 20
45 2024 I Problem 21
46 2024 I Problem 22
47 One-to-one mapping of the circle
48 2024 I Problem 23
49 The problem from MGTU
50 The trapezoid problem from MGTU

2025 I Problem 1

Let $I$ be the incenter of a triangle $\triangle ABC, D$ be an arbitrary point of sideline $AC,$ and $E, E'$ be the common points of the perpendicular from $D$ to the bisector $CI$ with $BC$ and $AI$ respectively. Define similarly the points $F, F'.$

Prove that $B, E, E', I, F,$ and $F'$ are concyclic.(Shvetzov)

Proof

$AD = AF', CD = CE.$ $\angle F'FI = \angle CFD = \angle ADF' - \angle ACI = (90^\circ - \angle CAI) - \angle ACI=$ $= \angle CBI = \angle EBI = \angle IBF' \implies IBFF'$ is concyclic.

Similarly $IBEE'$ is concyclic.

$IF' = IE, \angle F'BI = \angle FBI, BI$ is common in triangles $\triangle BIF$ and $\triangle BIE \implies$ $\angle BE'I + \angle BEI = 180^\circ \implies BEIF'$ is concyclic $\implies BEE'IF'F$ is concyclic.

2025 I Problem 2

Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points. (Kuznetsov)

Proof

Let a point symmetrical to $A$ with respect to $Z$ lie on the circle $\Omega = \odot BCD.$ We perform a homothety with center $A$ and coefficient $\frac {1}{2}.$ The image $\Omega$ circle $\omega$ contains the midpoints of the segments $AB, AC, AD$ and the point $Z.$

Therefore, the statement of the problem is equivalent to the fact that the four circles passing through the midpoints of the segments connecting each point with the three remaining ones had a common point.

Denote $P_{AB}$ the midpoint of segment $AB,$ similarly define $P_{AC}, P_{AD},...,$ $\omega_A = \odot P_{AB}P_{AC}P_{AD}, \omega_B = \odot P_{AB}P_{BC}P_{BD}, \omega_C = \odot P_{AC}P_{BC}P_{CD}, \omega_D = \odot P_{AD}P_{BD}P_{CD}.$

Let $Z = \omega_A \cap \omega_C \ne P_{AC}.$

We use the properties of the midlines and get $BD || P_{AB}P_{AD} || P_{BC}P_{CD},...$

WLOG, analize the case, shown in diagram. $\angle ABC = \angle P_{AD}P_{BD}P_{CD}, \angle ABD = \angle P_{AC}P_{BC}P_{CD}, \angle CBD = \angle P_{AC}P_{AB}P_{AD}.$ In $\omega _A$ we get $\angle P_{AC}ZP_{AD}=\angle P_{AC}P_{AB}P_{AD} = \angle CBD.$

In $\omega_C$ we get $\angle P_{AC}ZP_{CD} = 180^\circ - \angle P_{AC}P_{BC}P_{CD} = 180^\circ - \angle ABD.$

So $\angle P_{AD}ZP_{CD} = \angle P_{AC}ZP_{CD} - \angle P_{AC}ZP_{AD} = (180^\circ - \angle ABD)- \angle P_{AC}P_{AB}P_{AD} =$ $= 180^\circ - \angle ABD - \angle CBD = 180^\circ - \angle ABC = 180^\circ - \angle P_{AD}P_{BD}P_{CD} \implies$ $Z \in \omega_D.$ Similarly, $Z \in \omega_B. \blacksquare$

2025 I Problem 3

An excircle centered at $I_A$ touches the side $BC$ of a $\triangle ABC$ at point $D.$ Prove that the pedal circles of $D$ with respect to the triangles $ABI_A$ and $ACI_A$ are congruent. (Belsky)

Proof

Let the foots of perpendiculars from the point $D$ to $AI_A, AB, AC, BI_A,$ and $CI_A$ be $H, F, F', E,$ and $E',$ respectively, point $O-$ be the midpoint $AD, 2s$ be the semiperimeter of $\triangle ABC.$

$A -$ excircle touch the sides $AB$ and $AC$ at points $L$ and $L',$ respectively.

The points $A, F, D, H,$ and $F'$ are concyclic $(O$ is the center of this circle), so $\angle FOH = \angle F'OH = 2\angle BAI_A = \angle BAC.$

$DE = EL \implies OE$ is the midline of $\triangle DAL \implies OE ||AB, OE = AL/2 = s.$

Similarly, $OE' ||AC, OE' = AL'/2 = s \implies \angle EOE' = \angle BAC.$

The rotation centered at $O$ with angle $\angle BAC$ maps $\triangle FEH$ into $\triangle HE'F'$ therefore this triangles (and circumcircles) are congruent. $\blacksquare$

Proof 2

Denote $\angle BAC = 2 \alpha, \angle ABC = 2 \beta,$ $\angle ACB = 2 \gamma, \alpha + \beta + \gamma = 90^\circ.$

WLOG, $\beta \ge \gamma.$

Angles with vertex $I_A$ are $\angle BI_AD = 90^\circ - \angle CBI_A = \beta, \angle CI_AD = \gamma,$ $\angle BI_AA = 180^\circ - \angle BAI_A - \angle ABI_A = \gamma,$ $\angle AI_AD = \angle BI_AD - \angle AI_AB = \beta - \gamma.$

The points $I_A,E,H,D,$ and $E'$ are concyclic

$(I_AD$ is the diameter of this circle), so $\angle HED = \angle HI_AD = \angle AI_AD = \beta - \gamma.$

The points $B,E,D,$ and $F$ are concyclic

$(BD$ is the diameter of this circle), so $\angle FED = \angle FBD = \angle ABC = 2 \beta \implies$ $\angle FEH = \angle FED - \angle HED = \beta + \gamma = 90^\circ - \alpha.$

Similarly, $\angle F'E'H = 90^\circ - \alpha.$

The points $A, F, D, H,$ and $F'$ are concyclic

$(AD$ is the diameter of this circle), so $\angle FAH= \angle F'AH \implies FH = HF'.$

$R_{HEF} = \frac {HF}{\sin \angle FEH} = \frac {HF'}{\sin \angle F'E'H} = R_{HE'F'}.$

2025 I Problem 4

Let a triangle $\triangle ABC$ with the bisector $AL$ of $\angle A$ and point $X (XA \perp AL)$ be given. Let the line $\ell$ be the bisector of $AL, P = BX \cap \ell, Q = CX \cap \ell.$ Prove that points $A, X, P,$ and $Q$ are concyclic. (Shcherbatov)

Proof

Denote $D = AB \cap \ell, E = AC \cap \ell \implies DM = EM \implies ADLE$ is rhomb $\implies DL || AC.$

$\frac {BP}{BX} = \frac {BD}{BA} = \frac {BL}{BC} \implies PL || XC \implies \angle LPQ = \angle QXA.$

$\angle APM = \angle APQ = \angle LPM \implies A, X, P, Q$ are concyclic. $\blacksquare$

Proof 2

Denote $D = BC \cap AX, E = LX \cap \ell.$

We use the properties of bisectors and get: $\frac {BL}{LC} = \frac {AB}{AC} = \frac {DB}{DC},$ so $L$ is the harmonic conjugate of $D$ with respect to $C$ and $B$ and the cross-ratio $(D,B; L,C) = -1.$

Under projecting a straight line $BC$ to $\ell$ from point $X$ the cross-ratio is preserved, $C$ maps into $Q, L$ maps into $E, B$ maps into $P, D$ maps into point in infinity, so $\frac {\vec {PE}}{\vec {QE}} = -1 \implies E$ is the midpoint $PQ.$

$LE = EX \implies PLQX$ is parallelogram, so $QX = LP.$

$AP = LP = QX, PQ || AX \implies$

$APQX$ is the isosceles trapezoid, which is the cyclic quadrilateral. $\blacksquare$

2025 I Problem 5

Let $M$ be the midpoint of the cathetus $AC$ of a right-angled $\triangle ABC (\angle C = 90^\circ).$ The perpendicular from $M$ to the bisector of $\angle ABC$ meets $AB$ at point $N.$

Prove that the circumcircle of triangle $ANM$ touches the bisector of angle $\angle ABC.$ (Shvetsov)

Proof

Let the bisector of $\angle ABC$ meets circumcircle of $\triangle ABC$ at point $D, \angle ABD = \beta.$

$\angle ADB = 90^\circ,$ so circle $\omega$ with diameter $AD$ touches the bisector $AD.$ $D$ is the midpoint of arc $AC,$ so $DM \perp AC \implies M \in \omega.$

$MN \perp BD \implies MN || AD \implies \angle AMN = \angle DAM = \angle DAC = \angle DBC = \angle DBA = \beta.$

$\angle BAD = 90^\circ - \angle ABD = \angle NAD = 90^\circ - \beta,$ $\angle NMD = 90^\circ + \angle AMN = 90^\circ + \beta \implies N \in \omega. \blacksquare$

2025 I Problem 6

One bisector of a given triangle is parallel to one sideline of its Nagel triangle.

Prove that one of two remaining bisectors is parallel to another sideline of the Nagel triangle. (Emelyanov)

Proof

Let $\triangle ABC$ be the given triangle, $A'B'C'$ be the Nagel triangle $(A' \in BC), AF, BE$ be the bisectors, $a = |BC|, b = |AC|, c= |AB|, s = \frac{a+b+c}{2}.$

It is known that $AB' = A'B = s - c, AC' = A'C = s - b, CB' = C'B = s - a, AE = \frac {bc}{a+c}, BF = \frac {ac}{b+c}.$

WLOG, $B'C' || EB \Leftrightarrow \frac {AB'}{AE} = \frac{(s-c)(a+c)}{bc} = \frac {AC'}{AB} = \frac{s-b}{c} \Leftrightarrow$ $(s-c)(a+c) = (s-b)b \Leftrightarrow c^2 = a^2 + b^2.$

$c^2 = a^2 + b^2 \Leftrightarrow (s-c)(b+c) = (s-a)a \Leftrightarrow$

$\frac {A'B}{BF} = \frac{(s-c)(b+c)}{ac} = \frac {BC'}{AB} = \frac{s-a}{c} \Leftrightarrow A'C' || FA. \blacksquare$

Proof 2

Denote $\alpha, \beta, \gamma -$ semiangles of $\triangle ABC, I_A, I_C-$ the $A-, C-,$ excenters, $\omega$ is $C-$ excircle, $R$ is it’s radius, $D = BC \cup \omega_C, E = AC \cup \omega_C.$ $\angle BI_CC' = \beta, B'C = BC' = BD = s - a = R \tan \beta,$ $\angle BDC' = \frac {\angle ABC}{2} \implies$ points $B', C', D$ are collinear. $CD = s = R \cot \gamma, \angle CB'D = 180^\circ - 2 \gamma - \beta.$ We use the Sine Law for $\triangle B'CD$ and get: $\tan \beta \cdot \tan \gamma = \frac {s-a} {s} = \frac {\sin \beta}{\sin (2 \gamma + \beta)} \implies$ $\cot \gamma = \sin 2 \gamma + \cos 2\gamma \cdot \tan \beta.$ $\gamma = 45^\circ$ is the solution. If $\gamma \ne 45^\circ$ then $\tan \beta = \cot \gamma = \tan (90^\circ - \gamma).$

$\beta < 90^\circ, 90^\circ - \gamma < 90^\circ \implies \beta = 90^\circ - \gamma \implies \alpha = 90^\circ - \gamma - \beta = 0.$

It is impossible, so $\angle ACB = 90^\circ \implies I_AA' = CA' = AC' = AE.$

$I_AA' \perp BC, AC \perp BC \implies I_AA' || AE \implies I_AA'EA$ is parallelogram, so $I_AA||A'C'. \blacksquare$

2025 I Problem 7

Let $I, I_A$ be the incenter and the $A-$ excenter of a triangle $\triangle ABC, E, F$ be the touching points of the incircle with $AC, AB,$ respectively; $G$ be the common point of $BE$ and $CF.$ The perpendicular to $BC$ from $G$ meets $AI$ at point $J.$

Prove that points $E, F, I_A,$ and $J$ are concyclic. (Shcherbatov)

Proof

Denote $\angle BAC = 2\alpha, \angle AEJ = \varphi, a,b,c,$ and $s$ are sides of $\triangle ABC$ and it’s semiperimeter. $D$ is the touching points of the incircle with $BC,$ $L$ is the touching points of A-excircle with $AC.$ $J' = \odot EFI_A \cup AI.$

It is known that $G$ is the Gergonne point, $G \in AD.$ One can use formulas for barycentric coordinates of $G = (\frac{1}{s - a}:\frac{1}{s - b}:\frac{1}{s - c})$ or formulas of crossing segments $\frac{AE}{EC}= \frac {s - a}{s - c}$ and $\frac{AF}{FB}= \frac {s - a}{s - b}$ and get $\frac{DG}{AG} = \frac{(s - c)(s - b)}{a(s - a)} = \frac {s}{a} \tan ^2 \alpha.$ $ID || JG \implies \frac{IJ}{AJ} =\frac{DG}{AG} = \frac {s}{a} \tan ^2 \alpha.$

$\angle AJ'E + \angle IJ'E = 180^\circ, \angle IEA = 90^\circ,$ $\angle AEJ' = \varphi \implies \angle IEJ' = 90^\circ - \varphi.$ $\tan \varphi = \frac {LE}{LI_A} = \frac {s - (s-a)}{s \cdot \tan \alpha} =\frac{a}{s} \cot \alpha.$

We use the Sine Law for triangles $\triangle AJ'E$ and $\triangle IJ'E$ and get: $\frac {IJ'}{AJ'} = \frac {IE \cdot \sin IEJ'}{AE \cdot \sin AEJ'} = \frac {IE}{AE} \cdot \cot \varphi = \tan \alpha \cdot \cot \varphi = \frac{s}{a} \tan^2 \alpha.$

Therefore $J' = J. \blacksquare$

2025 I Problem 8

The diagonals of a cyclic quadrilateral $ABCD$ meet at point $P.$ Points $K$ and $L$ lie on $AC$ and $BD,$ respectively in such a way that $CK = AP$ and $DL = BP.$

Prove that the line joining the common points of circles $ALC$ and $BKD$ passes through the mass-center of $ABCD.$ (Konyshev)

Proof

Denote $\theta = \odot ABCD, \Omega = \odot ALC, \omega = \odot BKD,$ $R,Q = \omega \cap \Omega, N$ is the midpoint of $BD ($ and $PL), N'$ is the midpoint of $AC($ and $PK), G$ is the midpoint of $NN', M$ is the midpoint of $LK.$

It is clear that $G$ is the mass-center of $ABCD.$

We use properties of the crossing chords and get: $AP \cdot PC = BP \cdot PD (\theta) \implies Pow_{P, \omega} = Pow_{P, \Omega} \implies P \in PQ.$

$FL \cdot LE = BL \cdot LD = BP \cdot PD = AP \cdot PC = AK \cdot KC = FK \cdot KE \implies FL = KE.$ Similarly, $F'L = KE' \implies MF' \cdot MK = ML \cdot ME' \implies$ $Pow_{M, \omega} = Pow_{M, \Omega} \implies M \in PQ.$

$PM$ is the median of $\triangle PKL,$ so point $G$ lies on $PM$ which is the part of the radical axis of $\Omega$ and $\omega.$ $\blacksquare$

Proof 2

$AC$ is the radical axis of $\theta$ and $\Omega, BD$ is the radical axis of $\theta$ and $\omega \implies P = AC \cap BD$ is the radical center of $\theta, \omega,$ and $\Omega \implies P \in RQ (RQ$ is the radical axis of $\omega$ and $\Omega).$

It is known that the difference in the power of a point relative to a fixed pair of circles is a linear function of the coordinates of the point. Denote $F(X) = Pow_{X,\Omega} - Pow_{X,\omega}.$

Denote power of the point $P$ as $\mathcal{P} = AP \cdot PC = BP \cdot PD, F(P) = 0.$ $BP = LD \implies BL \cdot LD = \mathcal{P} = Pow_{L,\omega}, Pow_{L,\Omega} = 0 \implies F(L) = -\mathcal{P}.$

$AP = KC \implies AK \cdot KC = \mathcal{P} = Pow_{K,\Omega}, Pow_{K,\omega} = 0 \implies F(K) = \mathcal{P}.$ $4G = 2N + 2N' = 2P + L+ K \implies 4 F(G) = 2 F(P) + F(L) + F(K) = 0. \blacksquare$

2025 I Problem 9

The line $\ell$ passing through the orthocenter $H$ of a $\triangle ABC (BC > AB)$ and parallel to $AC$ meets $AB$ and $BC$ at points $D$ and $E,$ respectively. The line passing through the circumcenter of the triangle $O$ and parallel to the median $BM$ meets $\ell$ at point $F.$

Prove that the length of segment $HF$ is three times greater than the difference of $FE$ and $DH.$ (Mardanov)

Proof

Denote $G = HO \cup BM, N = DE \cup BM, DH = x,$ $NF = y, FE = z.$

$G$ is the centroid, so $HG = 2 GO.$ $GN || OF \implies HN = 2y.$ $DH + HN = NE = NF + EF \implies x + 2y = y + z \implies z = x+y.$ $HF = 3y, FE - DH = z - x = y \implies 3(FE - DH) = HF.$

2025 I Problem 10

An acute-angled triangle with one side equal to the altitude from the opposite vertex is cut from paper. Construct a point inside this triangle such that the square of the distance from it to one of the vertices equals the sum of the squares of distances to the remaining two vertices. No instruments are available, it is allowed only to fold the paper and to mark the common points of folding lines.(Evdokimov)

The Huzita–Justin axioms are a set of rules related to the mathematical principles of origami, describing the operations that can be made when folding a piece of paper. The axioms are as follows:

1. Given two distinct points $P_1$ and $P_2,$ there is a unique fold that passes through both of them.

2. Given two distinct points $P_1$ and $P_2,$ there is a unique fold that places $P_1$ onto $P_2.$

3. Given two lines $l_1$ and $l_2,$ there is a fold that places $l_1$ onto $l_2.$

4. Given a point $P$ and a line $l,$ there is a unique fold perpendicular to $l$ that passes through point $P.$

5. Given two points $P_1$ and $P_2$ and a line $\ell,$ there is a fold that places $P_1$ onto $\ell$ and passes through $P_2.$

6. Given two points $P_1$ and $P_2$ and two lines $l_1$ and $l_2,$ there is a fold that places $P_1$ onto $l_1$ and $P_2$ onto $l_2.$

7. Given point $P$ and two lines $l_1$ and $l_2,$ there is a fold that places $P$ onto $l_1$ and is perpendicular to $l_2.$

Solution

Let $AD, BE, CF$ be the heights of $\triangle ABC, AD = BC.$

Let $X \in BE$ be the point such $BX = BD.$ $AX^2 - CX^2 = AB^2 - BC^2 = AB^2 - AD^2 = BD^2 = BX^2.$ Similarly $AY^2 - BY^2 = CD^2 = CY^2.$

We construct heights $AD, BE, CF$ according the Axiom 4.

We construct bisector of angle $\angle CBF$ , fold places line $BC$ onto line $BF.$ (Axiom 3)

We construct point $X$ symmetric to $D$ with respect bisector of $\angle CBF. \blacksquare$

Solution 2

Let $Z \in AD$ be the point such $AZ^2 = CD^2 - BD^2 ($ WLOG $CD \ge BD).$ $CZ^2 - BZ^2 = CD^2 - BD^2 = AZ^2.$ We construct height $AD$ according the Axiom 4.

We construct point $C_1 \in BC, BC_1 = DC$ using midpoint of $BC.$

We construct point $K \in AD, BK = BC_1 = DC,$ a fold places $C_1$ onto $\ell = AD$ and passes through $B.$ (Axiom 5) $DK = \sqrt{CD^2 - BD^2} = AZ.$ We construct point $Z$ symmetric to $K$ with respect midpoint of $AD.\blacksquare$

2025 I Problem 11

A point $X$ is the origin of three rays such that the angle between any two of them equals $120^\circ .$ Let $\omega$ be an arbitrary circle with radius $R$ such that $X$ lies inside it, and $A, B, C$ be the common points of the rays with this circle.

Find $\max (XA + XB + XC).$ (Nilov)

Answer: 3R.

Solution

The Fermat–Torricelli point of a triangle with largest angle at most 120° is a point such that:

- the sum of the three distances from each of the three vertices of the triangle to the point is the smallest possible,

- the angles subtended by the rays to the vertieces of the triangle at Fermat–Torricelli point are all equal to 120°.

Therefore, $X$ is the Fermat–Torricelli point, $XA + XB + XC \le OA + OB + OC = 3R,$ where $O$ is the center of the circle. Equality is achieved when $O$ coincides with $X.\blacksquare$

Solution 2

Let $O$ be the circumcenter $\triangle ABC, OA' || XA, A' \in \omega.$ Define $B'$ and $C'$ similarly.

WLOG , $X$ lies in angle $\angle A'OC'.$ Let $D, F$ be the foots of perpendiculars from $X$ to $A'O$ and $C'O.$ Let $E$ be the foot of perpendiculars from $O$ to $BX.$

It is known that $|OD|+|OF| = |EX| = OX \cos \angle EXO.$ $AX + BX + CX < A'D + C'F + B'O + EX =$ $A'D + DO + C'F + FO + B'O =$ $= A'O + C'O + B'O = 3R. \blacksquare$ Solution 3

The triangle formed by the perpendiculars drawn at points $A, B, C$ to the rays $XA, XB, XC$ is regular.

The sum $AX + BX + CX$ is equal to the height of this regular triangle.

The sides of this regular triangle have common points with $\omega$ , therefore its height (and side) is maximum if $\omega$ is an inscribed circle.

This means the sum is maximum if $X$ is the center of the circle. $\blacksquare$

2025 I Problem 12

Let circles $\omega$ and $\Omega$ centered at $O$ and $O'$ be given, $M$ is the midpoint of $OO'.$ Let $X$ and $Y$ be arbitrary points on $\omega$ and $\Omega,$ respectively such that $MX = MY.$

Find the locus of the midpoints of segments $XY.$ (Shatunov)

Solution

Case 1

Let $O = O'.$ Then $M = O,$ so $\omega$ and $\Omega$ coincide. Any point inside this circle belong the locus.

Case 2

Circles $\omega$ and $\Omega$ have not common point. Denote $Z -$ midpoint of $XY, D = \omega \cup XY \ne X, D' = \Omega \cup XY \ne Y, P, P' -$ the midpoints of $XD$ and $YD'.$ $MX = MY, ZX = ZY \implies MZ \perp XY \implies MZ ||PO || P'O'.$ $MO = MO' \implies PZ = P'Z \implies PX = P'Y \implies DZ = D'Z.$ $Pow_{Z,\omega} = ZD \cdot ZX = ZD' \cdot ZY = Pow_{Z,\Omega}.$ Therefore $Z$ lies on the radical axes of $\omega$ and $\Omega.$

Let common external tangents crossed the radical axes at points $A$ and $A',$ common internal tangents crossed the radical axes at points $B$ and $B'.$

For given point $X$ there are two points $Y$ and $Y'$ $(MX = MY = MY'),$ corresponding to the condition. Respectively, there are two points $Z \in AB$ and $Z' \in A'B'.$ It seems obvious that for any point of these segments one can find points $X$ and $Y$ satisfying the condition. There was no need to prove this.

Case 3

Circles $\omega$ and $\Omega$ are crossing. Common external tangents crossed the radical axes at points $A$ and $A',$ and segment $AA'$ is the locus. $\blacksquare$

2025 I Problem 13

Each two opposite sides of a convex $2n-$ gon are parallel. (Two sides are opposite if one passes $n - 1$ other sides moving from one side to another along the borderline of the $2n-$ gon.) The pair of opposite sides is called regular if there exists a common perpendicular to them such that its endpoints lie on the sides and not on their extensions.

Which is the minimal possible number of regular pairs? (Frankin)

Answer: 1

Proof

1. We need to prove that zero regular pair is impossible.

Suppose, all pairs are irregular. Let $AB$ and $A'B'$ be a pair of sides with the minimum distance $(BD)$ between them, $BB' < AA', BC$ and $B'C'$ is the next irregular pare.

$2n-$ gon is convex, so acute $\angle DBC > \angle DBA$ and $BD$ cross $B'C'$ at point $E, BE < BD.$

Let $BH \perp BC \implies BH < BE < BD$ which contradicts the choice of the pair $AB, A'B'.$

2. We need to show that exist $2n-$ gon with only $1$ regular pair. WLOG, we do this for $n = 5.$

We use an obtuse angle $\angle BAC$ and divide $AB$ and $AC$ into $(n-1)$ equal segments as shown in diagram.

$BD = DX = XY = YA, AY' = Y'X' = X'G = GC,$ $E \in DY', XE||AC, F \in EX', YE ||AC.$

We construct the perpendicular to $FG$ through $G$ and construct the center $O$ of $2n-$ gon such that $OX || AC$ and $O$ lyes in the semiplane opposite to $AB.$

Points $B', D', E', F',$ and $G'$ are symmetrical to points $B, D, E, F,$ and $G$ with respect $O.$

Only sides $BG'$ and $B'G$ are the regular pair.

2025 I Problem 14

A point $D$ lies inside a triangle $ABC$ on the bisector of angle $B.$ Let $\omega_1$ and $\omega_2$ be the circles touching $AD$ and $CD$ at $D$ and passing through $B, P$ and $Q$ be the common points of $\omega_1$ and $\omega_2$ with the circumcircle $\Omega$ of $\triangle ABC$ (distinct from $B.$ ) Prove that the circumcircles of the triangles $\triangle PQD$ and $\triangle ACD$ are tangent. (L.Shatunov)

Proof

Denote $\theta = \odot PDQ, \Theta = \odot ADC, T = BD \cap \Omega \ne B,$ $X = TP \cap CD, Y = TQ \cap AD, Z = TP \cap AC.$

$PD$ is the chord of $\theta, AD$ is the chord of $\Theta.$

It seems evident that iff $\angle ADP = \angle PQD + \angle ACD$ then exist line through $D$ which is the common tangent to both of these circles.

$AD$ is the tangent to $\omega_1 \implies \angle ADP = \angle PBD.$ $\angle PBD = \angle PBT = \angle PQT = \angle PQY.$

So $\angle ADP = \angle PQY$ and quadrilateral $PDYQ$ is cyclic in $\theta.$

Similarly, $X \in \theta, XPDYQ$ is cyclic.

$2\angle AZP = \overset{\Large\frown} {AP} + \overset{\Large\frown} {CT} = \overset{\Large\frown} {AP} + \overset{\Large\frown} {AT} = 2 \angle PBT = 2 \angle ADP \implies A,P,D,Z$ are concyclic.

$\angle XYA = \angle XYA = \angle ZPD = \angle ZAD \implies XY || AC.$

Quadrilateral $PDYXQ$ is cyclic in $\theta \implies$ $\angle ADP = \angle PXY = \angle PXD + \angle YXD = \angle PQD + \angle ACD.\blacksquare$

Proof 2

Let's perform the inversion centered at $D$ with radius $BD.$ In inversion plane images of $\odot PQD$ and $\odot ACD$ need be parallel.

As a result, points $A, B, C, P,$ and $Q$ maps into $A', B, C', P',$ and $Q', \Omega$ maps into circle $\Omega' = \odot A'BC'P'Q',$ circle $\odot PBD$ maps into the line $BP'||AA'D,$ circle $\odot QBD$ maps into the line $BQ'||CDD'.$

By the properties of inversion, $\angle DA'B = \angle ABD = \angle CBD = \angle DC'B.$

$\angle A'BP' = \angle BA'D (BP' || A'D), \angle C'BQ' = DC'B (BQ' || C'D) \implies$ $\overset{\Large\frown} {A'P'} = \overset{\Large\frown} {C'Q'}$ in the circle $\Omega'.$

Therefore, the lines $A'C'||P'Q'.$ Their preimages $\odot PQD$ and $\odot ACD$ are tangent. $\blacksquare$

2025 I Problem 15

Let the point $C$ on the bisector of an acute angle with vertex $S$ be given. Let points $P$ and $Q$ be the foot from $C$ to the sidelines of the angle. The circle centered at $C$ with radius $PQ$ meets the sidelines at points $A$ and $B, (SA \ne SB).$ Denote $\omega$ the circle centered at $A$ and tangent to ray $SB, \Omega$ the circle centered at $B$ and tangent to ray $SA.$

Prove that $\omega$ and $\Omega$ are tangent. (Zaslavsky)

Proof

Denote $a = SC, \angle CSA = \angle CSB = \alpha \implies CP = CQ = a \sin \alpha,$ $SP = SQ = a \cos \alpha, PQ = 2 SP \sin \alpha = a \sin 2 \alpha.$ $AD = SA \sin 2 \alpha,BE = SB \sin 2 \alpha \implies AD + BE = (SA + SB) \sin 2 \alpha.$ $AC = BC, CP = CQ, \angle APC = \angle BQC = 90^\circ \implies$ $\triangle APC = \triangle BQC \implies AP = QB \implies SA + SB = SA + AP + SB - BQ = 2 SP.$ $\angle PAC = \angle CBQ \implies$ points $S,A,C,B$ are concyclic. $AB = 2 SC \cos \alpha \sin 2 \alpha = 2 SP \sin 2\alpha = AD + BE.$

2025 I Problem 16

The Feuerbach point of a scalene triangle lies on one of its bisectors. Prove that it bisects the segment between the corresponding vertex and the incenter. (Zaslavsky)

Proof

The Feuerbach hyperbola is a rectangular hyperbola passing through the vertices, orthocenter, incenter, Gergonne point, Nagel point and Schiffler point. The center of the hyperbola is the Feuerbach point. If vertex $A,$ incenter $I,$ and Feuerbach point $F$ are collinear, $A$ and $I$ need be on the different branches of hyperbola, therefore $A$ and $I$ are symmetrical with respect $F.$

Proof 2

Let given triangle be $\triangle ABC,$ the incenter $I,$ the inradius $r,$ the Feuerbach point $F, G$ is the point at $A-$ height such that $AG = r, N$ be the midpoint of $AI.$

It is known that points $F, G, N$ are collinear ( Feuerbach line). $\triangle ABC$ is a scalene triangle, so $G$ can not lie at $AI,$ and A-Feuerbach line not coincide with bisector $AI.$

The Feuerbach point lies on A-bisector therefore it coincide with $N.$

Proof 3

If the Feuerbach point of a scalene triangle lies on one of its bisectors, then the angle corresponding to this bisector is $60^\circ.$ Feuerbach point of a scalene triangle Denote $r$ the inradius and $AI$ - this bisector. Then $r = AI \sin \frac {60^\circ}{2} = \frac {AI}{2}, AF = AI - r = r.$ Proof 4

Let $F \in BI, K, L, M$ be the midpoints $AC, AB, BC,$ respectively. So the center $Q$ of nine-point circle $\omega = \odot KLM$ lies on $BFI.$

Let $D$ and $E$ be the foots from $Q$ to $\overline{BC}$ and $\overline{AB}, QE = QD.$ $QL = QM = \frac{R}{2},$ where $R$ is the circumradius of $\triangle ABC.$

$\triangle ABC$ is a scalene triangle, so $BM \ne BL.$ $\triangle DMQ = \triangle ELQ \implies \angle BLQ + \angle BMQ = 180^\circ.$

So quadrilateral $BLQM$ is cyclic. $\triangle BLM \sim \triangle BAC \implies \odot BLM$ has the radius $\frac{R}{2}$ and the center $Q$ of $\omega$ lies on $\odot BLM \implies$ $\angle LQM = 120^\circ \implies \angle ABC = 60^\circ \implies BF = BI - r = r.$

2025 I Problem 17

Let $O, I$ be the circumcenter and the incenter of an acute-angled scalene triangle $\triangle ABC, D, E, F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC, AB,$ respectively. Prove that if the orthocenter $H$ of $\triangle DEF$ lies on the circumcircle of $\triangle ABC,$ then it is symmetric to the midpoint of the arc $BC$ with respect to $OI.$ (Puchkov, Utkin)

Proof

Let $A', B', C'$ be the midpoints of $EF, DF, DE,$ respectively.

Let $I_A$ be A-excenter of $\triangle ABC$ and the circumcenter of $\Theta = \odot DEF,$ it’s radius $I_AE = 2r.$

Let $N$ be the center of $\theta \odot A'B'C'$ and nine-points center of $\triangle DEF \implies NA' = r.$

Points $H, N, I_A$ are collinear at Euler line of $\triangle DEF, I_AN = HN.$

$AF$ and $AE$ are tangents, $A'$ is the midpoint $EF$ so $A'$ is the inversion of point $A$ with respect $\Theta.$ Similarly, $B'$ and $C'$ so $\Omega = \odot ABC$ is the inversion $\theta$ with respect $\Theta.$ Therefore centers $O, N,$ and $I_A$ are collinear.

Homothety centered at $I_A$ with coefficient $k = \frac {R}{r}$ maps $\theta$ into $\Omega \implies I_AO = k \cdot I_AN = \frac {R}{r}I_AN = \frac {R}{2r}I_AH.$

If $H \in \Omega$ than $OH = OA = R.$ We use Euler formula $I_AO^2 = R^2 + 2R r_A$ and get $R \cdot I_AH = r I_AO \implies (R + 2r)(r-R)^2 = R^3,$ $4r = 3R \implies I_AO = 2R.$ The midpoint $I_AO$ lies on $\Omega, GH$ is the diameter of $\Omega.$

The midpoint of the arc $BC$ point $M$ is the midpoint of $II_A \implies GM ||OI, NM \perp GM, OI$ is the bisector of $HM.$

2025 I Problem 18

Let $ABCD$ be a quadrilateral such that the excircles $\omega$ and $\Omega$ of triangles $\triangle ABC$ and $\triangle BCD$ touching their sides $AB$ and $BD$ respectively touch the extension of $BC$ at the same point $P.$ The segment $AD$ meets $\Omega$ at point $Q,$ and the line $AD$ meets $\omega$ at $R$ and $S.$

Prove that one of angles $\angle RPQ$ and $\angle SPQ$ is right. (Kukharchuk)

Proof

$PC$ is the semiperimeter of $\triangle ABC$ and $\triangle BCD,$

$BC$ is the common side of these triangles, so $AB + AC = BD + CD.$

There is the circle $(\theta)$ inscribed in quadrilateral with diagonal $AD.$ https://en.wikipedia.org/wiki/Tangential_quadrilateral

Let $O,r$ and $O',R$ be the centers and the radii of $\omega$ and $\Omega, I$ be the Internal Similitude Center of circles $\Omega$ and $\omega,$ so $\frac{O'I}{OI} = \frac{R}{r}.$

WLOG, $R > r$ and lines $AB$ and $AC$ are the common internal tangents of $\omega$ and $\theta,$ so point $A$ is the Internal Similitude Center of circles $\theta$ and $\omega.$

Lines $CD$ and $BD$ are the common external tangents of $\Omega$ and $\theta,$ so point $D$ is the External Similitude Center of circles $\theta$ and $\Omega,$ therefore, points $A, D,$ and $I$ are collinear ( Monge's theorem for centers).

In accordance with Lemma, $\angle RPQ = 90^\circ.\blacksquare$

Lemma

Let circles $\omega$ and $\Omega$ centered at $O$ and $Q$ with radii $r$ and $R > r$ are internally tangent line $CD$ at point $P.$ Point $I$ lies at segment $OQ$ such that $\frac{OI}{QI} = \frac{r}{R} = k.$ Segment $AB, A \in \omega, A \ne P,$ $B \in \Omega, B \ne P$ contains $I.$

Prove that $\angle APB = 90^\circ.$

Proof

Let $E$ be the point on $\omega$ opposite $P.$ $OP = r, QO = R - r \implies OI = r \frac {R-r}{R+r}, PI = \frac{2rR}{R+r}, IE = \frac{2r^2}{R+r}.$

Homothety centered at $I$ with ratio $k = \frac{r}{R}$ sends point $P \in \Omega$ to point $E$ at $\omega.$ So radius $PQ$ maps to radius $EO$ which means $\Omega$ maps to $\omega.$

Point $B$ maps to point $A, \frac{BI}{AI} = \frac{R}{r} \implies \triangle AIO \sim \triangle BIQ.$ Denote $2\varphi = \angle PQB.$

Then $\angle BPC = \varphi, \angle AOP = 180^\circ - 2\varphi, \angle APD = 90^\circ - \varphi \implies$ $\angle APB = 180^\circ - \varphi - (90^\circ - \varphi) = 90^\circ.$

2025 I Problem 19

Let $I$ be the incenter of a triangle $ABC, A', B', C'$ be the orthocenters of the triangles $\triangle BIC, \triangle AIC, \triangle AIB,$

$M_A, M_B, M_C$ be the midpoints of $BC, CA, AB,$ and $S_A, S_B, S_C$ be the midpoints of $AA', BB', CC'.$

Prove that $M_AS_A, M_BS_B, M_CS_C$ are concurrent. (Kuznetsov)

Proof

$A'$ is the orthocenters of $\triangle BIC \implies A'C \perp BI.$

Similarly, $AC' \perp BI \implies A'C || AC'.$

$AS_A = A'S_A, CS_C = C'S_C \implies$ $A'C || AC' || S_AS_C.$ $S_AS_C$ is the midline of trapezium $AA'CC' \implies$ $M_B = AC \cap S_AS_C, M_B \in S_AS_C.$

Denote B-excenter of $\triangle ABC$ as $I_B, H$ and $H'$ the foots from $B'$ and $I_B$ to $AC.$

$HM_B = H'M_B \implies \triangle B'M_BH = \triangle I_BM_BH' \implies$ $B'M_B = I_BM_B.$ $I_B \in BI, BS_B = B'S_B, B'M_B = I_BM_B \implies$ $S_BM_B || BI.$ $BI \perp S_AS_C \implies S_BM_B \perp S_AS_C.$ Similarly, $S_AM_A \perp S_BS_C, S_CM_C \perp S_AS_B.$

Therefore $S_AM_A, S_BM_B, S_CM_C$ are the heights of $\triangle S_AS_BS_C$ which are crossed at the orthocenter of this triangle.

2025 I Problem 20

Let $H$ be the orthocenter of a triangle $ABC,$ and $M, N$ be the midpoints of segments $BC, AH,$ respectively. The perpendicular from $N$ to $MH$ meets $BC$ at point $A'.$ Points $B'$ and $C'$ are defined similarly.

Prove that points $A', B',$ and $C'$ are collinear. (Ivlev)

Proof

$MN$ is a diameter of the nine-point circle $\omega,$ so the projection of $N$ onto $MH$ point $E = MH \cap A'N$ lies on $\omega.$ The foot of perpendicular from $H$ to $BC$ point $D$ lies on $\omega.$

Therefore lines $NE$ and $MD$ crosses at point $A'$ lies on the polar $H$ with respect $\omega.$

Similarly $B',$ and $C'$ also lie on this polar. $\blacksquare$

Proof 2

Denote $O$ the circumcenter of $\triangle ABC,$ $\angle ABC = \beta, \angle ACB = \gamma,$ $\Omega = \odot ABC, \Theta = \odot BHC.$ WLOG $\beta \ge \gamma.$

It is known that raduis $AO || NM \implies$ $\angle DNM = \angle DAO = \angle OAB - \angle BAD =$ $(90^\circ - \gamma) - (90^\circ - \beta) = \beta - \gamma.$

$DN$ and $EM$ are the heights of $\triangle A'NM \implies A'H \perp NM \implies \angle MA'H = \beta - \gamma.$ $\angle BCH = 90^\circ - \beta, \angle CBH = 90^\circ - \gamma \implies$ $\angle A'HB = \angle CBH - \angle CA'H = 90^\circ - \beta = \angle BCH.$

Therefore, $A'H$ is the tangent $\Theta.$

Let $\theta$ be the circle centered at $H$ with zero radius. $A'H$ is the radical axis of $\theta$ and $\Theta, A'C$ is the radical axis of $\Omega$ and $\Theta,$ so $A'$ is the radical center of $\theta, \Omega,$ and $\Theta.$

$A'$ lies on radical axis of $\theta$ and $\Omega.$ Similarly, $B'$ and $C'$ lies on this radical axis. $\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

2025 I Problem 21

Let $P$ be a point inside a quadrilateral $ABCD$ such that $\angle APB + \angle CPD = 180^\circ.$ Points $A', B', C', D'$ are isogonally conjugated to $P$ with respect to the triangles $\triangle BCD, \triangle CDA, \triangle DAB, \triangle ABC,$ respectively.

Prove that the diagonals of the quadrilaterals $ABCD$ and $A'B'C'D'$ concur. (Galyapin)

Proof

Denote $F = AC \cap BD.$ By applying the Law of Sines on $\triangle ABF$ and $\triangle CBF$ we get: $\frac {AF}{\sin \angle ABD} = \frac {AB}{\sin \angle AFB}, \frac {CF}{\sin \angle CBF} = \frac {BC}{\sin \angle CFB}.$

$\angle AFB + \angle CFB = 180^\circ \implies \frac{AF}{CF} = \frac{AB \sin \angle ABD}{BC \sin \angle CBD}$

$\angle APB + \angle CPD = 180^\circ \implies$ there is the point $Q$ isogonally conjugated to $P$ with respect to quadrilateral $ABCD.$

The lines $A'C$ and $QC$ are isogonally conjugated to $PC$ with respect to $\angle BCD,$ so points $A', C,$ and $Q$ are collinear.

Similarly points $A, C',$ and $Q$ are collinear.

Point $Q$ isogonally conjugated to $P$ with respect to $\angle ABC \implies \angle ABP = \angle CBQ.$

Point $A'$ isogonally conjugated to $P$ with respect to $\angle CBD \implies \angle A'BC = \angle DBP.$

Adding last two equations we get $\angle ABP + \angle DBP = \angle ABD = \angle A'BC + \angle CBQ = \angle A'BQ.$ Similarly $\angle C'BQ = \angle CBD.$ $\angle A'BC =\angle ABQ - \angle CBQ = \angle ABD - \angle C'BQ = \angle ABC'.$

By applying the Law of Sines on $\triangle ABC'$ and $\triangle C'BQ$ we get: $\frac{C'Q}{AC'} = \frac{BQ \sin \angle C'BQ}{AB \sin \angle ABC'}.$

By applying the Law of Sines on $\triangle A'BQ$ and $\triangle A'BC$ we get: $\frac{A'C}{A'Q} = \frac{BC \sin \angle A'BC}{BQ \sin \angle A'BQ}.$

$\angle C'BQ = \angle CBD, \angle ABC' = \angle A'BC, \angle ABD = \angle A'BQ \implies$ $\frac{C'Q}{AC'} \cdot \frac{AF}{FC} \cdot \frac{A'C}{A'Q} = \frac{BQ \sin \angle C'BQ}{AB \sin \angle ABC'} \cdot \frac{AB \sin \angle ABD}{BC \sin \angle CBD} \cdot \frac{BC \sin \angle A'BC}{BQ \sin \angle A'BQ} = 1.$ According Menelaus theorem, points $A', C',$ and $F$ are collinear.

Similarly, points $B', D',$ and $F$ are collinear. $\blacksquare$

2025 I Problem 22

Let a circle $\omega$ and an ellipse with foci $F, F'$ inside it be given. Construct a chord $AB$ of the circle touching the ellipse such that $AFF'B$ is a cyclic quadrilateral.

Proof

Denote $D$ and $E$ the crosspoints of $\omega$ and line $FF'.$

Let $S \in FF'$ be the point with propertyes $SD \cdot SE = SF \cdot SF'.$ To construct this point one can construct any circle through the points $F$ and $F'$ and intersecting $\omega.$ The desired point is the intersection of the radical axis of these circles (common chord line) and the line $FF'$ .

Let $\Omega$ be the circumscribed circle of given ellipse.

Let $S'$ be the inverse of a point $S$ with respect to $\Omega.$

Let $C$ be the point of the ellipse such that $CS' \perp FF', A$ and $B$ be the crosspoints of $\omega$ and $CS.$

Then $CS$ is tangent to the ellipse and $AFF'B$ is a cyclic quadrilateral. (See Ellipse and tangent)

2025 I Problem 23

Let us say that a subset points $M$ of the plane contains a hole if there exists a disc not contained in $M,$ but contained inside some polygon with the boundary lying in $M.$ Can the plane be presented as a union of $N$ convex sets such that the union of any $N - 1$ from them contains a hole? (Spivak)

Solution

Let $N = 7.$ The first element of the desired set is a regular hexagon $ABCDEF,$ and the remaining six are half-planes adjacent to its edges external for hexagon.

It is clear that the last six has a hole which is a hexagon, and for any other group the hole is a regular triangle adjacent to the hexagon $(AFG,$ for half-plane $AF).$

2025 I Problem 24

The insphere of a tetrahedron $ABCD$ touches the faces $ABC, BCD, CDA, DAB$ at points $D', A', B', C',$ respectively. Denote by $[X]$ the area of the triangle $X.$

Prove that there exists a triangle with sidelengths $\sqrt{[AC'B] \cdot [CB'D]}, \sqrt{[AB'C] \cdot [BC'D]}, \sqrt{[AC'D] \cdot [BA'C]}.$

Proof

Let us consider the net of the tetrahedron $ABCD$ onto a plane $ABC.$ There is three points $D$ belonging faces $ABD, ACD,$ and $BCD.$ Denote $\angle AD'B = \alpha,$ $\angle BD'C = \beta,\angle AD'C = \gamma, \angle CA'D = x.$

Tangents from each vertex are equals $AB' = AC' = AD' = a, BA' = BC' = BD' = b,$ $CA' = CB' = CD' = c, DA' = DB' = DC' = d,$ $\angle AC'B = \alpha, \angle BA'C = \beta, \angle AB'C = \gamma,\angle CB'D = x.$ $\angle AB'D = 360^\circ - \angle AB'C - \angle CB'D =$ $= 360^\circ - \gamma - x = \angle AC'D.$ $\angle BC'D = 360^\circ - \angle AC'D - \angle AC'B =$ $= 360^\circ - (360^\circ - \gamma - x) - \alpha = \gamma + x - \alpha.$ $\angle BA'D = 360^\circ - \angle BA'C - \angle CA'D =$ $= 360^\circ - \beta - x = \angle BC'D = \gamma + x - \alpha.$ $2x = 360^\circ - \beta - \gamma + \alpha = 2 \alpha \implies$ $x = \alpha \implies \angle AB'D = \beta, \angle BA'D = \gamma.$ $\sqrt{[AC'B] \cdot [CB'D]} = \frac {1}{2} \sqrt{(AC' \cdot BC' \cdot \sin \alpha) \cdot (CB' \cdot DB' \cdot \sin \alpha)} = \frac {1}{2} \sqrt{abcd} \sin \alpha.$

Similarly $\sqrt{[AC'D] \cdot [BA'C]} = \frac {1}{2} \sqrt{abcd} \sin \beta, \sqrt{[AB'C] \cdot [BC'D]} = \frac {1}{2} \sqrt{abcd} \sin \gamma.$

Therefore three sidelength are proportional to $\sin \alpha, \sin \beta, \sin \gamma.$

Let us place three points $K,L,M$ in the unit circle, the angles between which are equal to $2 \pi - 2 \alpha , 2 \pi - 2 \beta , 2 \pi - 2 \gamma.$ Then the lengths of the chords are equal to $2\sin (\pi - \alpha) = 2\sin \alpha,...$

So the required triangle is constructed. $\blacksquare$

Proof

Proof

2024 II Problem 6

A point $P$ lies on one of medians of triangle $\triangle ABC$ in such a way that $\angle PAB = \angle PBC = \angle PCA.$ Prove that there exists a point $Q$ on another median such that $\angle QBA = \angle QCB = \angle QAC.$ (A.Zaslavsky)

Proof

1. Denote $a = BC, b = AC, c = AB.$ It is known that barycentric coordinates are $P = (a^2 b^2 : b^2 c^2 : c^2 a^2), P \in AD \implies b^2 c^2 = c^2 a^2 \implies AC = BC.$

2. Denote $\omega = \odot APB, \Omega = \odot APC.$

$\angle ABP = \angle PAC \implies AC$ is tangent to $\omega.$

$\angle PAC = \angle PCB \implies BC$ is tangent $\Omega.$

$PC$ is the radical axes of $\omega$ and $\Omega,$ the power of a point $D$ with respect to a circle $\Omega$ is $CD^2$ so the power of a point $D$ with respect to a circle $\omega$ is $CD^2.$

$BD^2 = CD^2,$ so $BD$ is tangent to $\omega \implies \angle ABC = \angle CAB \implies AC = BC.$

$AC = BC,$ so point $Q$ symmetrical to $P$ with respect to the $C-$ median satisfies the conditions. $\blacksquare$

2024 II Problem 2(8)

Let $M$ be the midpoint of side $AB$ of an acute-angled triangle $ABC,$ and $P$ be the projection of the orthocenter $H$ to the bisector of angle $C.$ Prove that $MP$ bisects the segment $CH.$ (L.Emelyanov)

Solution

Denote $D$ - the midpoint of $CH, A',B',$ and $C'$ the foots of the heights, $\angle A = 2 \alpha, \angle B = 2 \beta, \angle C = 2 \gamma, \omega$ be the Euler circle $A'DB'C'M.$

$\Omega$ is the circle $\odot CA'B'$ with the diameter $CH.$ $HP \perp CP \implies P \in \Omega.$ $AA' \perp BA' \implies A'M = BM \implies \angle BA'M = 2 \beta.$ $\triangle ABC \sim \triangle A'BC' \implies \angle BA'C' = 2 \alpha \implies$ $\angle MA'C' = 2|\alpha - \beta| = \angle MDC'.$ $\angle ACC' = 90^\circ - 2 \alpha, \angle ACP = \gamma,$ $\angle PCC' = |\angle ACC' - \angle ACP| = |90^\circ - 2 \alpha - \gamma| = | \alpha + \beta + \gamma - 2 \alpha - \gamma| = | \alpha - \beta|.$ $PD = CD \implies \angle PDH = 2 \angle PCD = 2|\alpha - \beta|.$ $\angle PDH = \angle MDH \implies$ points $M,P,$ and $D$ are collinear.

2024 II Problem 4(8)

A square with sidelength $1$ is cut from the paper. Construct a segment with length $\frac{1}{2024}$ using at most $13$ folds. No instruments are available, it is allowed only to fold the paper and to mark the common points of folding lines. (M.Evdokimov)

Solution

Main idea: $\frac{1}{2024} = \frac{1}{4} \cdot \left [ \frac{1}{22} - \frac{1}{23} \right ].$ $\frac {a}{b} = \frac{x}{x+c} \implies x = \frac {a \cdot c}{b - a}.$ $EG = EA - GA = c \cdot \left [ \frac {1}{b/a - 1} - \frac {1}{(b+FB)/a - 1} \right ].$ Let $c = \frac{1}{2}, a = \frac{1}{64}, b = \frac{45}{64}, FB = \frac{2}{64}.$ $EG = \frac{1}{2} \cdot \left [ \frac {1}{45 - 1} - \frac {1}{47 - 1} \right] = \frac{1}{2024}.$ We perform $1$ horizontal fold of the sheet. We get line $AD (AC = \frac{1}{2}).$ We perform

$6$ vertical folds of the sheet. We get $62$ vertical lines at a distance of $\frac{1}{64}$ from each other.

Point $F$ is the lower left corner of the sheet, point $B$ is the lower point of the second vertical line, point $C$ is the lower point of the $47^{th}$ line, point $D$ is the point at the intersection of the horizontal line and the $46^{th}$ vertical line.

Points $E$ and $G$ are at the intersection of the lines $BD$ and $FD$ and the $47^{th}$ vertical line.

2024 II Problem 3(9)

Let $(P,P')$ and $(Q,Q')$ be two pairs of points isogonally conjugated with respect to a triangle $ABC,$ and $R$ be the common point of lines $PQ$ and $P'Q'.$ Prove that the pedal circles of points $P,Q,$ and $R$ are coaxial. (L.Shatunov, V.Shelomovskii)

Solution

1. Let $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$ Then circle centered at the midpoint $PP'$ is the common pedal circle of points $P$ and $P'.$ ( Circumcircle of pedal triangles) So center $O$ is the midpoint $PP'$ and center $U$ is the midpoint $QQ'.$

2. Denote $R' = PQ' \cap P'Q.$ Then $R'$ is the isogonal conjugate of a point $R$ with respect to $\triangle ABC.$ So center $V$ is the midpoint $RR'.$ ( Two pares of isogonally conjugate points)

3. The Gauss line (or Gauss–Newton line) is the line joining the midpoints of the three diagonals of a complete quadrilateral $PQ'P'Q$ (Gauss line).So points $U,O,$ and $V$ are collinear as was to be proven. $\blacksquare$

2024 II Problem 4(9)

For which $n > 0$ it is possible to mark several different points and several different circles on the plane in such a way that:

- exactly $n$ marked circles pass through each marked point;

- exactly $n$ marked points lie on each marked circle;

- the center of each marked circle is marked? (P.Puchkov)

Solution

Case $n = 1.$ Circles centered at $A$ and $B$ with radii $R = |\vec {AB}|.$

Case $n = 2, \vec {AC} = \vec {BC'}, |\vec {AB} = \vec {AC}|, \vec {AC}$ is not paralel to $\vec {AB}.$

Four circles are centered at points $A, B, C,$ and $C'.$ Each radius is equal $R.$

Case $n = 3, \vec {CD} = \vec {AD''} = \vec {C'D'} = \vec {BD'''}, \vec {CD}$ is not paralel to $\vec {AB}$ or $\vec {AC}, |\vec {AB}| = \vec {CD}|$

Eight circles centered at $A, B, C, C', D, D',D''$ and $D'''$ have radii $R.$

Case $n = 4, \vec {DE} = \vec {D'E'} = ...,|\vec {DE}| = R,..$

Answer For all $n.$

2024 II Problem 5(9)

Let $ABC$ be an isosceles triangle $(AC = BC), O$ be its circumcenter, $H$ be the orthocenter, and $P$ be a point inside the triangle such that $\angle APH = \angle BPO = 90^\circ.$

Prove that $\angle PAC = \angle PBA = \angle PCB.$ (A.Zaslavsky)

Proof

Denote $E -$ the midpoint $BC, M-$ the midpoint $AB, D -$ the foot from $B$ to $\overline{AC}, \alpha = \angle ABC, \omega = \odot ADHM, \Omega = \odot BEOM, \Theta = \odot BHC, \theta = \odot AB$ tangent to $\overline{BC}.$

$AH = BH, BO = CO , \angle BAH = 90^\circ - \alpha = \angle BCO \implies \triangle ABH \sim \triangle BCO.$ There is a spiral similarity $T$ centered at point $X$ that maps $\triangle ABH$ into $\triangle BCO.$

The coefficient of similarity $k = \frac {BC}{AB} = \frac {1}{2 \cos \alpha},$ rotation angle equal $180^\circ - \alpha.$ $B = T(A), B = AB \cap BC \implies X \in \theta.$ $C = T(B), B = AB \cap BC \implies X \in \Theta$ $(\angle BCH = \angle ABH,$ so $AB$ is tangent to $\Theta).$ Basic information $E = T(M), B = AM \cap BE \implies X \in \Omega.$ $AB = T(CA), CO = T(BH), D = AC \cap BH, M = AB \cap CO \implies M = T(D).$ $A = AD \cap BM \implies X \in \omega.$ $P = \theta \cap \Theta \implies P = X.$ $P = T(P) \implies \triangle BPC = T(\triangle APB) \implies \triangle BPC \sim \triangle APB \implies \angle PAB = \angle PCB.$ $\triangle OBP = T(\triangle HAP) \implies \triangle OBP \sim \triangle HAP \implies \angle OBP = \angle HAP .$ $\angle PAC = \alpha - \angle HAP - \angle BAH = \alpha - \angle OBP - \angle OBC = \angle PBA.\blacksquare$ $\triangle ADM \sim \triangle ABC \implies \angle ADM = \alpha = \angle APM, \angle EPM = 180^\circ - \alpha \implies$ Points $A,P,$ and $E$ are collinear, so $P \in A-$ median of $\triangle ABC.$

$\angle PAB = \angle PCB, \angle ABC = \angle BAC \implies PC$ is $B-$ symmedian of $\triangle ABC.$

$\angle APH = 90^\circ \implies P$ is $A-$ Humpty point.

2024 II Problem 7(9)

Let triangle $\triangle ABC$ and point $P$ on the side $BC$ be given. Let $P'$ be such point on the side $BC$ that $BP = P'C.$ The cross points of segments $AP$ and $AP'$ with the incircle $\omega$ of $\triangle ABC$ form a convex quadrilateral $EFE'F'.$

Find the locus of crosspoints of diagonals $EFE'F'.$ (D.Brodsky)

Solution 1. Particular case of Fixed point .

2. Denote $p_a = \frac{b+c-a}{2}, p_b = \frac{a-b+c}{2}, p_c = \frac{a+b-c}{2}, m = \frac {CP}{BP}, \mu = \frac {PF'}{AF'}.$ $PD = \frac {m p_b - p_c} {m+1}, AD' = p_a \implies$ $x(x+y) = p_a^2, z(z+y) = PD^2, x+y+z = AP, \mu = \frac {z}{x+y}.$ We perform simple transformations and get: $\mu^2 p_a^2 - \mu (AP^2 - p_a^2 - PD^2) + PD^2 = 0.$ We use Stewart's theorem and get: $AP^2 = \frac {AC^2}{1+m} + \frac {m AB^2}{1+m} - \frac {m BC^2}{(1+m)^2} \implies$ $(\mu p_a)^2 - 2 (\mu p_a) \frac {m p_b + p_c}{m+1} + \frac {(m p_b - p_c)^2} {(m+1)^2} = 0.$ $\mu = \frac {(\sqrt{m p_b} \pm \sqrt{p_c})^2}{(m+1) p_a}.$ Similarly $\nu = \frac {\eta + \zeta}{\xi}= \frac {(\sqrt{m p_c} \pm \sqrt{p_b})^2}{(m+1) p_a}.$ Therefore $\frac {\nu + \mu}{2} = \frac {a}{2 p_a}$ not depends from $m.$

Let $M$ be the midpoint of $BC, AM$ is the median of $\triangle ABC$ and $\triangle PAP'.$

The line $FF'$ cross the median of $\triangle PAP'$ at point $G'$ such that $\frac {MG'}{G'A} = \frac {\nu + \mu}{2} = \frac {p_b + p_c}{2 p_a} = \frac {a}{2 p_a}.$

So point $G'$ is fixed and this point lyes on $EE' \implies G = G'$ .

Therefore the locus of crosspoints of diagonals $EFE'F'$ is point $G.$

Corollary

Let line $MQ||D'D'', Q \in AC$ . Then $\frac {D''Q}{AD''} = \frac {a}{2 p_a} \implies 2 D''Q = BC.$

2024 II Problem 7(10)

Let $ABC$ be a triangle with $\angle A = 60^\circ, AD, BE,$ and $CF$ be its bisectors, $P, Q$ be the projections of $A$ to $EF$ and $BC$ respectively, and $R$ be the second common point of the circle $\omega = \odot DEF$ with $AD.$

Prove that points $P, Q, R$ are collinear. (K.Belsky)

Proof

Denote $a = BC, b = AC, c = AB, I -$ the incenter of $\triangle ABC,$ $L=FE \cap AD, R' -$ the midpoint of $AI.$

It is known ( Division of bisector) that $\frac {AI}{DI} = \frac {b+c}{a}, \frac {DI}{IL} = 1+ \frac {2a}{b+c} \implies \frac{1}{IL} = \frac{1}{DI} + \frac{2}{AI} \implies$ $2 DI (\frac{AI}{2} - IL) =AI \cdot IL \implies 2 DI \cdot LR' = AI \cdot IL.$

$DL \cdot LR' = (DI + IL) \cdot LR' = \frac {AI \cdot IL}{2} + IL \cdot (\frac{AI}{2} - IL) = AI \cdot IL - IL^2 = AL \cdot IL.$ $\angle BAC = 60^\circ \implies \angle BIC = \angle FIE = 120^\circ \implies AEIF$ is cyclic.

Therefore $AL \cdot IL = LE \cdot LF = DL \cdot LR' \implies EDFR'$ is cyclic $\implies R = R'.$

Let $AG \perp AD, G \in BC, H = AP \cap BC, Q' = RP \cap BC.$

It is known that points $F, E,$ and $G$ are collinear, $\angle AGB = \frac{|\angle ACB - \angle ABC|}{2}, \angle AGE = \angle BGE.$

$AP \perp EF, AQ \perp BG \implies AG$ is the diameter of $\odot APQG \implies$ $\overset{\Large\frown} {QP} = \overset{\Large\frown} {AP} \implies \angle DAP = \angle QAP, AP = PH.$ $AR = RI, AP = PH \implies RP || IH \implies \frac {DH}{HQ'} = \frac {DI}{IR} = \frac {2a}{b+c}.$ $AH$ is the bisector of $\angle DAQ \implies \frac {DH}{HQ} = \frac {AD}{AQ}.$

Bisector $AD = \frac {2 b c \cos 30^\circ}{b+c}.$

Altitude $AQ = \frac {2[ABC]}{BC}= \frac {bc \sin 60^\circ}{a} \implies \frac {AD}{AQ} = \frac {2a}{b+c} = \frac {DH}{HQ} = \frac {DH}{HQ'} \implies Q = Q'. \blacksquare$

Note that the point $R$ is a Feuerbach point of $\triangle ABC$ since both the inscribed circle and the Euler circle pass through it.

2024 I Problem 2

Three distinct collinear points are given. Construct the isosceles triangles such that these points are their circumcenter, incenter and excenter (in some order).

Solution

Let $M$ be the midpoint of the segment connecting the incenter and excenter. It is known that point $M$ belong the circumcircle. Construction is possible if a circle with diameter IE (incenter – excenter) intersects a circle with radius OM (circumcenter – M). Situation when $E$ between $I$ and $O$ is impossible.

Denote points $A, B, C$ such that $B \in AC$ and $AB \le BC.$

Suppose point $A$ is circumcenter, so $B$ is incenter. $M$ is midpoint BC. The vertices of the desired triangle are located at the intersection of a circle with center $A$ and radius $AM$ with $\omega$ and a line $AB.$

Suppose point $C$ is circumcenter, so $B$ is incenter. $M$ is midpoint $AB.$ The vertices of the desired triangle are located at the intersection of a circle with center $A$ and radius $AM$ with $\omega$ and a line $AB.$

Suppose point $B$ is circumcenter, so $A$ is incenter. $M$ is midpoint $AB.$ Suppose $3 AB < BC.$ The vertices of the desired triangle are located at the intersection of a circle with center $B$ and radius $BM$ with $\omega$ and a line $AB.$

If $3 AB \ge BC$ there is not desired triangle.

2024 I Problem 8

Let $ABCD$ be a quadrilateral with $\angle B = \angle D$ and $AD = CD.$

The incircle of $\triangle ABC$ touches the sides $BC$ and $AB$ at points $E$ and $F$ respectively.

The midpoints of segments $AC, BD, AE,$ and $CF$ are points $M,X,Y,Z.$

Prove that points $M,X,Y,Z.$ are concyclic.

Solution

$ZM || AB, YM || BC \implies$ $\angle YMZ = \angle ABC = \angle ADC = \alpha.$ $2 \vec {XY} = \vec {DA} + \vec {BE}, 2 \vec {XZ} = \vec {DC} + \vec {BF}.$ $\vec {DC}$ is the rotation of $\vec {DA}$ around a point $D$ through an angle $\alpha.$

$\vec {BF}$ is the rotation of $\vec {BE}$ around a point $B$ through an angle $\alpha.$

So $\vec {XZ}$ is the rotation of $\vec {XE}$ around a point $X$ through an angle $\alpha.$

2024 I Problem 9

Let $ABCD (AD || BC$ be a trapezoid circumscribed around a circle $\omega,$ centered at $O$ which touches the sides $AB, BC, CD,$ and $AD$ at points $P, Q, R, S,$ respectively.

The line passing trough $P$ and parallel to the bases of trapezoid meets $QR$ at point $X.$

Prove that $AB, QS,$ and $DX$ concur.

Solution

Solution 1. $AP = AS \implies \angle AOS = \frac {\overset{\Large\frown} {PS}}{2} = \angle PQS \implies PQ||AO.$

$OD \perp OC, QR \perp OC \implies OD || QX. AD ||PX \implies$

$E = AP \cap QS$ is the center of similarity of triangles $\triangle PQX$ and $\triangle AOD.$

Solution 2. $\triangle ODS \sim \triangle QXG \implies \frac {SD}{GX} = \frac {SO}{GQ} = \frac {1}{2} \cdot \frac {SQ}{GQ} = \frac {1}{2} \cdot \frac {AB}{PB}.$

Denote $EA = x, AP = AS = y, BP = BQ = z.$

$\triangle AES \sim BEQ \implies \frac {x}{x+y+z}= \frac {y}{z} \implies xz = xy + y^2 + yz \implies 2xz = xy + y^2 + yz + xz \implies$

$\frac {x}{x+y}= \frac {y+z}{2z} \implies \frac {AS}{AG} = \frac {1}{2} \cdot \frac {AB}{PB} \implies \triangle ESD \sim \triangle EGX \implies E \in DX.$

2024 I Problem 12

The bisectors $AE, CD$ of a $\triangle ABC$ with $\angle B = 60^\circ$ meet at point $I.$

The circumcircles of triangles $ABC, DIE$ meet at point $P.$

Prove that the line $PI$ bisects the side $AC.$

Proof

Denote $M$ the midpoint $AC, \omega = \odot DIE,$ $\varphi = \angle CIM, \phi = \angle DIP, f(x) = \frac {\sin x}{\sin(120^\circ -x)}.$ $\angle AIC = 90^\circ + \frac {\angle ABC}{2} = 120^\circ = 180^\circ - \angle ABC \implies B \in \omega.$ In triangles $\triangle CIM$ and $\triangle AIM$ , by applying the law of sines, we get $\frac {IM}{\sin \angle ACI} = \frac {CM}{\sin \varphi}, \frac {IM}{\sin \angle CAI} = \frac {AM}{\sin (120^\circ -\varphi)} \implies f(\varphi) = \frac {\sin \varphi} {\sin (120^\circ -\varphi)} = \frac {\sin \angle ACI}{\sin \angle CAI}.$

We use the formulas for circle $\omega$ and get $\frac {PD}{\sin \angle PID} = \frac {PE}{\sin \angle PIE} \implies f(\phi) = \frac {\sin \phi} {\sin (120^\circ -\phi)} = \frac {PD}{PE}.$

$\angle BDI = \angle AEC \implies \angle ADC = 180^\circ - \angle AEC.$ In triangles $\triangle ADC$ and $\triangle AEC$ , by applying the law of sines, we get $\frac {AD}{\sin \angle ACD} = \frac {AC}{\sin \angle ADC} = \frac {AC}{\sin \angle AEC} \implies \frac {AD}{CE} = \frac {\sin \angle ACI}{\sin \angle CAI}.$

$\angle BEP = \angle BDP, \angle BCP = \angle BAP \implies \triangle APD \sim \triangle CPE \implies \frac {AD}{CE} = \frac {PD}{PE}.$

Therefore $f(\phi) = f(\varphi).$ The function $f$ increases monotonically on the interval $(0, \frac {2 \pi}{3}).$

This means $\phi = \varphi$ and points $P,I,$ and $M$ are collinear.

2024 I Problem 14

The incircle $\omega$ of a right-angled triangle $\triangle ABC$ touches the circumcircle $\theta$ of its medial triangle at point $F.$ Let $OE$ be the tangent to $\omega$ from the midpoint $O$ of the hypothenuse $AB$ distinct from $AB.$ Prove that $CE = CF.$

Proof

Let $\Omega$ and $I$ be the circumcircle and the incenter of $\triangle ABC, D = \omega \cap AB.$

Let $Q$ be nine-point center of $\triangle ABC, G$ be the point at $OB$ such that $OG = DO, K \in DI, KQ ||AB.$

Denote $r = ID, R = AO,a = BC, b = AC,c= AB, \beta = \angle ABC.$

$\triangle ABC$ is the right-angled triangle, so $Q$ is the midpoint $CO,$

$2R = c, r = \frac {a+b-c}{2},FQ = \frac{R}{2}.$ Let $h(X)$ be the result of the homothety of the point $X$ centered in $C$ with the coefficient $2.$ Then $h(\theta) = \Omega, F' = h(F) \in \Omega, O = h(Q) \implies FQ || F'O.$ $DO = EO = GO \implies \angle DEG = 90^\circ.$ $\angle DOE = 2 \angle DOI = 2 \angle EGD \implies \angle DOI = \angle EGD, IO || EG.$ WLOG, $a > b \implies DO = \frac {a-b}{2} = \tan IOD = \frac{r}{DO} = \frac{a+b-c}{a-b} = \frac{\cos\beta + \sin \beta -1}{\cos\beta - \sin \beta}.$

Let $H$ be the foot from $C$ to $\overline{AB}$ . $CH = a \sin \beta = c \sin \beta \cos \beta, BH = a \cos \beta, BG = AD = \frac{c+b-a}{2}.$ $\tan \angle CGH = \frac {CH}{BH - BG} = \frac{2 \sin \beta \cos \beta}{2 \cos^2 \beta +\cos \beta - \sin \beta -1} =$ $= \frac{(\sin \beta + \cos \beta -1)(\sin \beta + \cos \beta +1)}{(\cos \beta - \sin \beta)(\sin \beta + \cos \beta +1)} = \frac{\cos\beta + \sin \beta -1}{\cos\beta - \sin \beta} = \tan IOD.$ Therefore points $C,E,$ and $G$ are collinear. $\psi = \angle BCG = \angle AGC - \angle ABC \implies$ $\tan \psi = \frac {\tan \angle AGC - \tan \angle ABC}{1 + \tan \angle AGC \cdot \tan \angle ABC} = \frac {c - a}{c-b}.$ $\sin 2\psi = \frac {2\tan \psi}{1 + \tan^2 \psi} = \frac {2(c-a)(c-b)}{c(3c - 2b - 2a)},$ $\angle AOF' = \angle KQI, \sin \angle KQI = \frac{CH/2 -r}{QF - r}.$ $4c(QF - r) = c(c -2a -2b +2c) = c(3c -2a -2b),$ $2c(CH - 2r) = 2(ab -ac -bc +c^2) = 2(c - a)(c - b),$ $\sin \angle KQI = \frac {2(c-a)(c-b)}{c(3c - 2b - 2a)}= \sin 2 \psi \implies \angle KQI = 2 \psi.$ $\angle ACF' = \frac {\angle AOF'}{2} = \frac {\angle KQI}{2} = \psi = \angle BCG \implies \angle FCI = \angle ECI \implies CF = CE.$

2024 I Problem 15

The difference of two angles of a triangle is greater than $90^\circ.$ Prove that the ratio of its circumradius and inradius is greater than $4.$

Proof

Suppose, $\angle BAC = \alpha = \angle ACB + 90^\circ = \gamma + 90^\circ.$

Let $\Omega = \odot ABC, C'$ be the point on $\Omega$ opposite $C, B'$ be the midpoint of arc $\overset{\Large\frown} {AC}.$ Then $\angle CAC' = 90^\circ \implies \angle BAC' = \angle ACB = \gamma = \angle BCC' \implies$ $\overset{\Large\frown} {AB} = \overset{\Large\frown} {BC'} \implies BB' || CC'.$ $BO = CO \implies \angle CBO = \angle OCB = \angle ACB = \gamma.$ $\angle ABC = \beta = 180^\circ - \gamma - \alpha = 90^\circ - 2 \gamma.$ $\angle OBB' = \frac {\beta}{2} + \gamma = 45^\circ.$ $BO = OB' = R \implies \angle BB'O = 45^\circ, \angle BOB' = 90^\circ.$ Incenter $I$ triangle $\triangle ABC$ lies on $BB',$ therefore $OI \ge \frac {R}{\sqrt {2}}.$

We use the Euler law $OI^2 = R^2 - 2 Rr \implies \frac {2r}{R} = 1 - \frac {OI^2}{R^2} \le \frac{1}{2} \implies \frac{R}{r} \ge 4.$

If $R = 4r$ then $OI \perp BB' \implies \frac {BI}{IB'} = 1 \implies \frac {BC + AB}{AC} = 2, BI = \frac{R}{\sqrt{2}}.$ $\sin CBI = \frac {r}{BI} = \frac{r \sqrt{2}}{R} = \frac{1}{2\sqrt{2}} \implies \sin \beta = \frac{\sqrt{7}}{4} \implies$ $AC = 2R \sin \beta = R \frac {\sqrt{7}}{2},AB = AC (1 - \frac{1}{\sqrt {7}}), BC = AC (1 + \frac{1}{\sqrt {7}}).$

If $\angle BAC' > \angle ACB \implies \frac {OI}{R}$ increases so $\frac {r}{R}$ decreases.

2024 I Problem 16

Let $AA', BB',$ and $CC'$ be the bisectors of a triangle $\triangle ABC.$

The segments $BB'$ and $A'C'$ meet at point $D.$ Let $E$ be the projection of $D$ to $AC.$

Points $P$ and $Q$ on the sides $AB$ and $BC,$ respectively, are such that $EP = PD, EQ = QD.$

Prove that $\angle PDB' = \angle EDQ.$

Proof

$\triangle PDQ = \triangle PEQ (DQ = EQ, DP = PF, PQ$ is the common side) $\implies$

$PQ \perp DE, F = PQ \cap DE$ is the midpoint $DE \implies$

$G = BB' \cap PQ$ is the midpoint of $DB'.$ $\frac{BG}{BB'} =\frac {BQ}{BC} = \frac {BP}{BA} = \frac {BD}{BI}.$ (see Division of bisector for details.)

So $DQ || CC', PD || AA'.$ Denote $\angle ACC' = \angle BCC' = \gamma, \angle A'AC = \alpha, B'BC = \beta.$ $\angle PDB' = \angle AIB' = \angle BB'C - \angle IAC = 180^\circ - \beta - 2 \gamma - \alpha = 90^\circ - \gamma.$ $\angle QDE = 90^\circ - \angle DQP = 90^\circ - \gamma = \angle PDB'.$

Another solution see 2024_Sharygin_olimpiad_Problem_16

2024 I Problem 17

Let $\triangle ABC$ be not isosceles triangle, $\omega$ be its incircle.

Let $D, E,$ and $F$ be the points at which the incircle of $\triangle ABC$ touches the sides $BC, CA,$ and $AB,$ respectively.

Let $K$ be the point on ray $EF$ such that $EK = AB.$

Let $L$ be the point on ray $FE$ such that $FL = AC.$

The circumcircles of $\triangle BFK$ and $\triangle CEL$ intersect $\omega$ again at $Q$ and $P,$ respectively.

Prove that $BQ, CP,$ and $AD$ are concurrent.

Proof

$\frac {KE} {FL} = \frac {AB}{AC},$ so points $Q,P,$ and $G = FE \cap BC$ are collinear (see Symmetry and incircle for details).

Therefore lines $BQ, CP,$ and $AD$ are concurrent (see Symmetry and incircle A for details.)

2024 I Problem 18

Let $AH, BH', CH''$ be the altitudes of an acute-angled triangle $ABC, I_A$ be its excenter corresponding to $A, I'_A$ be the reflection of $I_A$ about the line $AH.$ Points $I'_B, I'_C$ are defined similarly. Prove that the lines $HI'_A, H'I'_B, H''I'_C$ concur.

Proof

Denote $I$ the incenter of $\triangle ABC.$ Points $A, I, I_A$ are collinear. We will prove that $I \in HI'_A.$ Denote $D \in BC, ID \perp BC, D' \in I'_AI_A, ID' \perp BC,$ $E = BC \cap AI_A, F \in BC, I_AF \perp BC, AH = h_A,$ $ID = r, I_AF = r_A, BC = a, s$ - semiperimeter. $\frac {HD}{HF} = \frac {AI}{AI_A} = \frac {h_A - r}{h_A + r}.$ The area $[ABC] = r \cdot s = r_A (s - a) = \frac {a h_A}{2} \implies$ $\frac {1}{r} = \frac {1}{r_A} + \frac {2}{h_A} \implies \frac {h_A - r}{h_A+ r} = \frac {r_A}{r}.$ $\frac {I'_AD'}{HD} = \frac {HD + HF}{HD} = 1 + \frac {HF}{HD} = 1 + \frac {r_A}{r}= \frac {r_A + r}{r} \implies$ Points $I, H, I'_a$ are collinear, so the lines $HI'_A, H'I'_B, H''I'_C$ concur at the point $I.$

2024 I Problem 19

A triangle $ABC,$ its circumcircle $\Omega$ , and its incenter $I$ are drawn on the plane.

Construct the circumcenter $O$ of $\triangle ABC$ using only a ruler.

Solution

We successively construct:

- the midpoint $D = BI \cap \Omega$ of the arc $AB,$

- the midpoint $E = CI \cap \Omega$ of the arc $AC,$

- the polar $H'H''$ of point $H \in DE,$

- the polar $G'G''$ of point $G \in DE,$

- the polar $F = H'H'' \cap G'G''$ of the line $DE,$

- the tangent $FD || AC$ to $\Omega,$

- the tangent $FE || AB$ to $\Omega,$

- the trapezium $ACDF,$

- the point $K = AF \cap CD,$

- the point $L = AD \cap CF,$

- the midpoint $M = AC \cap KL$ of the segment $AB,$

- the midpoint $M'$ of the segment $AC,$

- the diameter $DM$ of $\Omega,$

- the diameter $EM'$ of $\Omega,$

- the circumcenter $O = DM \cap EM'.$

2024 I Problem 20

Let a triangle $ABC,$ points $D$ and $E \in BD$ be given, $F = AD \cap CE.$ Points $D', E'$ and $F'$ are the isogonal conjugate of the points $D, E,$ and $F,$ respectively, with respect to $\triangle ABC.$

Denote $R$ and $R'$ the circumradii of triangles $\triangle DEF$ and $\triangle D'E'F',$ respectively.

Prove that $\frac {[DEF]}{R} = \frac {[D'E'F']}{R'},$ where $[DEF]$ is the area of $\triangle DEF.$

Proof

Denote $\angle BAC = \alpha, \angle ABC = \beta, \angle ACB = \gamma,$ $\angle EDF = \Theta, \angle E'D'F' = \theta, \angle DEF = \Psi, \angle D'E'F' = \psi,$ $\angle BAD' = \angle CAF = \varphi_A, \angle CBD' = \angle ABD = \varphi_B, \angle BCE' = \angle ACE = \varphi_C.$ It is easy to prove that $\frac {[DEF]}{R} = \frac {[D'E'F']}{R'}$ is equivalent to $DE \cdot \sin \Theta \cdot \sin \Psi = D'E' \cdot \sin \theta \cdot \sin \psi.$ $\Theta = \alpha - \varphi_A + \varphi_B, \theta = \beta - \varphi_B + \varphi_A, \Psi = \beta - \varphi_B + \varphi_C, \psi = \gamma - \varphi_C + \varphi_B \implies$ $\theta = 180^\circ - \gamma - \Theta, \psi = 180^\circ - \alpha - \Psi.$ By applying the law of sines, we get $\frac {BE}{\sin \varphi_C} = \frac {BC}{\sin \Psi}, \frac {BD}{\sin (\alpha - \varphi_A)} = \frac {AB}{\sin \Theta}, \frac {BC}{\sin \alpha} = \frac {AB}{\sin \gamma}.$ $ED = BD - BE \implies DE \cdot \sin \Theta \cdot \sin \Psi = \frac {AB}{\sin \gamma} \left ( \sin \gamma \cdot \sin (\alpha - \varphi_A) \cdot \sin \Psi - \sin \alpha \cdot \sin \varphi_C \cdot \sin \Theta \right )$

$\frac {BE'}{\sin (\gamma - \varphi_C)} = \frac {BC}{\sin \psi}, \frac {BD'}{\sin \varphi_A} = \frac {AB}{\sin \theta}.$ $D'E' = BE' - BD' \implies D'E' \cdot \sin \theta \cdot \sin \psi = \frac {AB}{\sin \gamma} (\sin \alpha \cdot \sin (\gamma - \varphi_C) \cdot \sin \theta - \sin \gamma \cdot \sin \varphi_A \cdot \sin \psi ).$ We need to prove that $\sin \gamma \cdot \sin (\alpha - \varphi_A) \cdot \sin \Psi - \sin \alpha \cdot \sin \varphi_C \cdot \sin \Theta = \sin \alpha \cdot \sin (\gamma - \varphi_C) \cdot \sin \theta - \sin \gamma \cdot \sin \varphi_A \cdot \sin \psi$ We make the transformations: $\sin \alpha \left[ \sin \varphi_C \cdot \sin \Theta + \sin (\gamma - \varphi_C) \cdot \sin (\gamma + \Theta) \right] = \sin \gamma \left[ \sin (\alpha - \varphi_A) \cdot \sin \Psi + \sin \varphi_A \cdot \sin (\alpha + \Psi) \right]$

$\sin \alpha \left[ \sin \varphi_C \cdot \sin \Theta + (\sin \gamma \cos \varphi_C - \cos \gamma \sin \varphi_C) \cdot (\sin \gamma \cos \Theta + \cos \gamma \sin \Theta \right] =$ $= \sin \gamma \left[ \sin \alpha \cos \varphi_A \cdot \sin \Psi - \cos \alpha \sin \varphi_A \cdot \sin \Psi + \sin \varphi_A \cdot \sin \alpha \cos \Psi + \sin \varphi_A \cdot \cos \alpha \sin \Psi \right]$ $\sin \alpha \left[ \sin \varphi_C \cdot \sin \Theta + \sin^2 \gamma \cos \varphi_C \cos \Theta - \cos ^2 \gamma \sin \varphi_C \cdot \sin \Theta + \cos \gamma \sin \Theta + \sin \gamma \cos \gamma (\cos \varphi_C \sin \Theta - \sin \varphi_C \cos \Theta) \right] =$ $= \sin \gamma \cdot \sin \alpha \left[\cos \varphi_A \cdot \sin \Psi + \sin \varphi_A \cdot \cos \Psi \right]$

$\sin \alpha \left[ \sin^2 \gamma \cdot \cos (\Theta - \varphi_C) + \sin \gamma \cdot \cos \gamma \cdot \sin (\Theta - \varphi_C)\right] = \sin \gamma \cdot \sin \alpha \cdot \sin (\Psi + \varphi_A)$

$\sin \alpha \cdot \sin \gamma \cdot \sin (\gamma + \Theta - \varphi_C) = \sin \gamma \cdot \sin \alpha \cdot \sin (\beta + \varphi_A - \varphi_B + \varphi_C).$ The last statement is obvious.

2024 I Problem 21

A chord $PQ$ of the circumcircle of a triangle $ABC$ meets the sides $BC, AC$ at points $A', B',$ respectively. The tangents to the circumcircle at $A$ and $B$ meet at point $X,$ and the tangents at points $P$ and $Q$ meets at point $Y.$ The line $XY$ meets $AB$ at point $C'.$

Prove that the lines $AA', BB',$ and $CC'$ concur.

Proof

WLOG, $P \in \overset{\Large\frown} {AC}.$ Denote $\Omega = \odot ABC, Z = BB' \cap AA', D = AQ \cap BP.$

Point $D$ is inside $\Omega.$

We use Pascal’s theorem for quadrilateral $APQB$ and get $D \in XY.$

We use projective transformation which maps $\Omega$ to a circle and that maps the point $D$ to its center.

From this point we use the same letters for the results of mapping. Therefore the segments $AQ$ and $BP$ are the diameters of $\Omega, C'D \in XY || AP \implies C'$ is the midpoint $AB.$

$AB|| PQ \implies AB || B'A' \implies C' \in CZ \implies$

preimage $Z$ lies on preimage $CC'.\blacksquare$

2024 I Problem 22

A segment $AB$ is given. Let $C$ be an arbitrary point of the perpendicular bisector to $AB,$ $O$ be the point on the circumcircle of $\triangle ABC$ opposite to $C,$ and an ellipse centered at $O$ touche $AB, BC, CA.$

Find the locus of touching points $P$ of the ellipse with the line $BC.$

Solution

Denote $M$ the midpoint $AB, D$ the point on the line $CO, DO = MO, \alpha = \angle CBM, b = OM.$

$\angle CBO = 90^\circ \implies$ $\angle COB = \alpha, MB = b \tan \alpha,$ $CB = \frac {b \sin \alpha}{\cos^2 \alpha}, CO = \frac {b} {\cos^2 \alpha},$ $CD = b \left (1 + \frac {1} {\cos^2 \alpha} \right ).$ In order to find the ordinate of point $P,$ we perform an affine transformation (compression along axis $AB)$ which will transform the ellipse $MPD$ into a circle with diameter $MD.$ The tangent of the $CP$ maps into the tangent of the $CE, E = \odot CBO \cap \odot MD, PF \perp CO.$ $\angle OEF = \angle ECO \implies OF = OE \sin \angle OEF = OE \sin \angle ECO = b \cos^2 \alpha.$ $CP = \frac {CF}{\sin \alpha} = \frac {b}{\sin \alpha}\left ( \frac {1} {\cos^2 \alpha} - \cos^2 \alpha \right ) = b \sin \alpha \left ( \frac {1}{\cos^2 \alpha } + 1 \right).$ $\frac {CP}{CD} = \sin \alpha , \angle PCD = 90^\circ - \alpha \implies \angle CPD = 90^\circ.$ $BP = CP - CB = b \sin \alpha.$ Denote $Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.$

So point $Q$ is the fixed point ( $P$ not depends from angle $\alpha, \angle BPQ = 90^\circ ).$

Therefore point $P$ lies on the circle with diameter $BQ$ (except points $B$ and $Q.)$

One-to-one mapping of the circle

Let a circle $\Omega,$ two fixed points $A$ and $B$ on it and a point $C$ inside it be given. Then there is a one-to-one mapping of the circle $\Omega$ onto itself, based on the following two theorems.

1. Let a circle $\Omega,$ two fixed points $A$ and $B$ on $\Omega,$ and a point $C$ inside $\Omega$ be given.

Let an arbitrary point $Q \in \Omega$ be given.

Let $A' = AC \cap \Omega, A' \ne A, B' = BC \cap \Omega, B' \ne B. S$ is the midpoint of the arc $A'B', D = A'Q \cap BS, E = B'Q \cap AS.$

Denote $\omega = \odot AEC, \omega' = \odot BCD, P = \omega \cap \omega'.$ Prove that $P \in \Omega.$

2. Let a circle $\Omega,$ two fixed points $A$ and $B$ on $\Omega,$ and a point $C$ inside $\Omega$ be given.

Let an arbitrary point $P \in \Omega$ be given.

Let $A' = AC \cap \Omega, A' \ne A, B' = BC \cap \Omega, B' \ne B. S$ is the midpoint of the arc $A'B'.$

Denote $\omega = \odot ACP, \omega' = \odot BCP, D = BS \cap \omega' , E = \omega \cap AS.$

Denote $Q = A'D \cap B'E.$ Prove that $Q \in \Omega.$

Proof

$1. C = AA' \cap BB', D = A'Q \cap BS, E = B'Q \cap AS \implies$

Points $D,C,E$ are collinear. $\angle APC = \angle SEC, \angle BPC = \angle SDC.$ $\angle APC + \angle BPC + \angle ASC = \angle SEC + \angle SDC + \angle DSE = 180 ^\circ \implies P \in \Omega.$

2. Points $D, C,$ and $E$ are collinear (see Claim in 2024, Problem 23).

We use Pascal's theorem for points $A,B',S,A',B$ and crosspoints $C,D,E$ and get $Q \in \Omega.$

2024 I Problem 23

A point $P$ moves along a circle $\Omega.$ Let $A$ and $B$ be fixed points of $\Omega,$ and $C$ be an arbitrary point inside $\Omega.$

The common external tangents to the circumcircles of triangles $\triangle APC$ and $\triangle BCP$ meet at point $Q.$

Prove that all points $Q$ lie on two fixed lines.

Solution

Denote $A' = AC \cap \Omega, B' = BC \cap \Omega, \omega = \odot APC,$ $\omega' = \odot BPC, \theta = \odot ACB', \theta' = \odot BCA'.$

$O$ is the circumcenter of $\triangle APC, O'$ is the circumcenter of $\triangle BPC.$

Let $K$ and $L$ be the midpoints of the arcs $\overset{\Large\frown}{CB'}$ of $\theta,D = AL \cap \omega.$

Let $K'$ and $L'$ be the midpoints of the arcs $\overset{\Large\frown}{CA'}$ of $\theta', D' = BL' \cap \omega'.$

These points not depends from position of point $P.$

Suppose, $P \in \overset{\Large\frown} {B'ABA'} ($ see diagram). $\angle A'BC = 2 \alpha = \angle B'AC \implies \angle D'BC = \angle DAC = \alpha \implies \angle DOC = \angle D'O'C = 2 \alpha.$ $O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.$ Let $F= CD \cup OO' \implies \frac {FO}{FO'} = \frac {OC}{O'D'} \implies Q = F.$

$\angle LCB' = \alpha = \angle B'BL' \implies LC || L'B.$ Similarly, $AL || CL' \implies \triangle DLC \sim \triangle CL'D' \implies \frac {LC}{L'D'} = \frac {DC}{CD'} = \frac {OC}{O'D'}.$

Let $F' = LL' \cap DD' \implies \frac {F'C}{F'D'} = \frac {LC}{L'D'} = \frac {OC}{O'D'}= \frac {FC}{FD'} \implies F' = F.$

Therefore $Q \in LL'.$ Similarly, if $P \in \overset{\Large\frown} {B'A'}$ then $Q \in KK'.$

Claim

Points $D, C,$ and $D'$ are collinear.

Proof

$S$ is the midpoint of arc $\overset{\Large\frown}{A'B'} \implies \angle SAC = \angle SBC.$ Denote $\angle CAP = \alpha, \angle CBP = \beta, \angle SAC = \angle SBC = \varphi.$ $D \in \omega \implies \angle PDC = \alpha, \angle PCD = \pi - \alpha - \varphi.$ $D' \in \omega' \implies \angle PD'C = \beta, \angle PCD' = \pi - \beta - \varphi.$ $S \in \Omega \implies \angle SAP + \angle SBP = \alpha + \beta + 2 \varphi = \pi.$ Therefore $\angle PCD + \angle PCD' = \pi \implies$ points $D, C,$ and $D'$ are collinear.

The problem from MGTU

The lateral face of the regular triangular pyramid $SABC$ is inclined to the plane of the base $ABC$ at an angle of $\alpha = \arctan \frac{3}{4}.$ Points $D, E, F$ are the midpoints of the sides of the $\triangle ABC.$ Triangle $\triangle DEF$ is the lower base of a right prism. The edges of the upper base of the prism intersect the lateral edges of the pyramid $SABC$ at points $K, L, N.$ The area of the total surface of the polyhedron with vertices $D, E, F, K, L, N$ is equal to $53 \sqrt{3}.$ Find the side of $\triangle ABC.$

Solution

Denote $AB = BC = AC = a, O$ is the center of $\triangle ABC, G = DF \cap AO, GK \perp ABC, K \in AS, M = SD \cap KN.$ $OD = \frac{ CD}{3} = \frac {a \sqrt {3}}{6}, SO = OD \cdot \tan \alpha = \frac {a \sqrt {3}}{8}.$ $AG = \frac {AE}{2}, AO = \frac{2AE}{3} \implies \frac {KG}{SO} = \frac{AG}{AO} = \frac{3}{4} \implies KG = \frac {3 a \sqrt{3}}{32} = h.$ $MD = \frac{KG}{\sin \alpha} = \frac{5}{3} KG = \frac{5h}{3}.$ The area of the total surface of the polyhedron with vertices $D, E, F, K, L, N$ is $Area = [DEF] + [KLN] + 3[DFK]+ 3[DKN].$ $[ABC] = \frac {a^2 \sqrt{3}}{4} = 4[DEF] =4 s \implies s = \frac {a^2 \sqrt{3}}{16}.$ $\frac{KN}{AB} = \frac {SK}{SA} = \frac{GO}{AO} = \frac {1}{4} \implies [KLN] =\frac{[ABC]}{16} = \frac {s}{4} \implies [DEF] + [KLN] = \frac {5s}{4}.$ $[DFK] = \frac{KG \cdot DF}{2} = \frac{ah}{4}, ah = \frac{3}{2} s.$ $[DKN] = \frac{MD \cdot KN}{2} = \frac{5ah}{24} \implies 3[DFK]+ 3[DKN] = \frac{11ah}{8} = \frac{33s}{16}.$ $Area = \frac {5s}{4} + \frac{33s}{16} = \frac{53s}{16} = 53 \sqrt{3} \implies s = 16 \sqrt {3} = \frac {a^2 \sqrt{3}}{16} \implies a = 16.$

The trapezoid problem from MGTU

Points $M$ and $N$ are the midpoints of bases $AD$ and $BC$ of trapezoid $ABCD.$

Denote $\alpha$ the angle between lines $MN$ and $AC, \cos \alpha = \frac{11}{16}.$

Find the area of trapezoid $ABCD$ if $MN = 2, BD = 6.$

Solution

$AF||MN||CG \implies AG = CF = AM+NC \implies FB=DG \implies$ $[ABCD] = [AFCG] = AC \cdot CG \cdot \sin \alpha = 2 CH \cdot CG \cdot \sin \alpha.$ By applying the Law of Cosines on $\triangle CHG, HG = \frac {BD}{2} = 3,$ we get $CH^2 + CG^2- 2 CH \cdot CG \cos \alpha = GH^2 \implies CH^2 - \frac {11}{4} CH - 5 = 0 \implies CH = 4.$ $\sin ^2 \alpha = 1 - \cos^2 \alpha = \frac {135}{16^2} \implies [ABCD] = 2 \cdot 4 \cdot 2 \frac {\sqrt{135}}{16} = \sqrt{135}.$

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Sharygin Olympiads, the best

Contents

2025 I Problem 1

2025 I Problem 2

2025 I Problem 3

2025 I Problem 4

2025 I Problem 5

2025 I Problem 6

2025 I Problem 7

2025 I Problem 8

2025 I Problem 9

2025 I Problem 10

2025 I Problem 11

2025 I Problem 12

2025 I Problem 13

2025 I Problem 14

2025 I Problem 15

2025 I Problem 16

2025 I Problem 17

2025 I Problem 18

2025 I Problem 19

2025 I Problem 20

2025 I Problem 21

2025 I Problem 22

2025 I Problem 23

2025 I Problem 24

Proof

2024 II Problem 6

2024 II Problem 2(8)

2024 II Problem 4(8)

2024 II Problem 3(9)

2024 II Problem 4(9)

2024 II Problem 5(9)

2024 II Problem 7(9)

2024 II Problem 7(10)

2024 I Problem 2

2024 I Problem 8

2024 I Problem 9

2024 I Problem 12

2024 I Problem 14

2024 I Problem 15

2024 I Problem 16

2024 I Problem 17

2024 I Problem 18

2024 I Problem 19

2024 I Problem 20

2024 I Problem 21

2024 I Problem 22

One-to-one mapping of the circle

2024 I Problem 23

The problem from MGTU

The trapezoid problem from MGTU