Geo challenge on finding simple ways to solve it

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Geo challenge on finding simple ways to solve it

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Bulgaria Spring Mathematical Competition 2025 9.2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1238 posts

#1 h Mar 30, 2025, 12:35 PM • 1 Y

Y by ehuseyinyigit

Let $ABC$ be an acute scalene triangle inscribed in a circle $\Gamma$ . The angle bisector of $\angle BAC$ intersects $BC$ at $L$ and $\Gamma$ at $S$ . The point $M$ is the midpoint of $AL$ . Let $AD$ be the altitude in $\triangle ABC$ , and the circumcircle of $\triangle DSL$ intersects $\Gamma$ again at $P$ . Let $N$ be the midpoint of $BC$ , and let $K$ be the reflection of $D$ with respect to $N$ . Prove that the triangles $\triangle MPS$ and $\triangle ADK$ are similar.

This post has been edited 1 time. Last edited by Assassino9931, Mar 30, 2025, 1:09 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1475 posts

#2 h Mar 30, 2025, 2:10 PM • 1 Y

Y by Funcshun840

Let $S'$ midpoint of arc $BAC$ on $\Gamma$ , let $AA'CB$ isosceles trapezoid, let $AA_1$ diameter of $\Gamma$ and let $T$ point on $BC$ such that $AT$ is tangent to $\Gamma$ . And finally let $S'L \cap \Gamma=E$ which by ratio Lemma it happens that $AE$ is symedian.
Claim 1: $T,P,A_1$ are colinear.
Proof: From Reim's theorem we have $P,D,A'$ colinear and thus stacking ratio lemmas:
$\frac{BP}{PC} \cdot \frac{BA_1}{A_1C}=\frac{BD}{DC} \cdot \left(\frac{CA'}{A'B} \right)^2 \cdot \frac{BK}{KC}=\left( \frac{BA}{AC} \right)^2=\frac{BT}{TC}$ Happens to finish (notice $A',K,A_1$ colinear from reflecting was used).
To finish: Now just note that $ADKA'$ is a rectangle so $\measuredangle MSP=\measuredangle AA'D=\measuredangle AKD$ but also using Claim 1 and projecting cross ratios:
$-1=(A, E; P, A_1) \overset{S'}{=} (A, L; S'P \cap AL, \infty_{AL}) \implies P,M,S' \; \text{colinear!}$ and from that we get $\measuredangle SPM=90=\measuredangle KDA$ thus we are done :cool:

:cool:

.

This post has been edited 1 time. Last edited by MathLuis, Mar 30, 2025, 2:10 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1238 posts

#3 h Mar 30, 2025, 5:15 PM • 1 Y

Y by ehuseyinyigit

Here is my (not too complicated) solution, though some contestants claimed that there are even easier approaches (i.e. not involving the midpoint of arc $BAC$ ), though I don't know their details.

Without loss of generality, we assume $AB > AC$ . Using standard angle notations for the triangle, we have $\angle LMD = 180^\circ - 2\angle DLM = 180^\circ - 2(\beta + \frac{\alpha}{2}) = \gamma - \beta$ . Also, $\angle APD = \angle APS - \angle DPS = \gamma + \frac{\alpha}{2} - \angle DLA = \gamma - \beta$ , which means quadrilateral $AMDP$ is cyclic. From here, we find $\angle MPS = \angle MPD + \angle DPS = \angle MAD + \angle DLA = 90^\circ$ .

Let $MP$ intersect $\Gamma$ at point $T$ . Thus, $T$ is the midpoint of arc $BAC$ on $\Gamma$ because $\angle SPT = 90^\circ$ . We have $AD \parallel TN \perp BC$ , so $TN$ intersects $AK$ at its midpoint $W$ (from the midsegment in $\triangle ADK$ ). Therefore, $\angle TAS = 90^\circ$ since $ST$ is a diameter of $\Gamma$ , and $\angle TWM = \angle TNB = 90^\circ$ due to the parallelism of $MW$ and $DK$ . Hence, $ATWM$ is cyclic, leading to $\angle AKD = \angle AWM = \angle ATM = \angle ATP = \angle ASP = \angle MSP$ , which concludes the proof.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6 posts

#4 h Mar 30, 2025, 8:12 PM • 2 Y

Y by Assassino9931, bo18

Very straightforward geo problem.

Without loss of generality, we assume $AB > AC$ .

It's clear that $ML=MA=MD$ . Let $P'$ be the second point of intersection of the circumcircles of $\triangle DSL$ and $\triangle MAD$ .
$\angle SP'A = \angle SP'D + \angle AP'D = \angle MLD + \angle DML = \angle MDC = \angle ALB = 180^\circ - \beta - \frac{\alpha}{2}$
$\angle SBA = \beta + \frac{\alpha}{2}$
So $P'$ lies on $\Gamma$ , meaning $P'=P$
$\angle SPM = \angle SPD + \angle MPD = \angle DLA + \angle DAL = 90^\circ$

Let the line $SP$ cross the line $BC$ at $X$ . So the points $B, K, L, D, C, X$ lie on the line $BC$ in that order.
$XC = a, CD = BK = b, DL = c, LK = d$
We have $XC.XB = XS.XP = XD.XL$ by Power of a point, which simplifies to $a(a+2b+c+d) = (a+b)(a+b+c)$ or $ad = b^2 + bc$ or $ad+bd+cd = b^2+bc+bd+cd$ or $d(a+b+c) = (b+c)(b+d)$ so $KL.LX = BL.LC$ .
But by Power of a point $BL.LC = AL.LS$ so $AL.LS = KL.LX$ which implies that $AKLX$ is cyclic. Now $\angle AKD = \angle ASP$ and we conclude by showing that the two angles in $\triangle MPS$ and $\triangle ADK$ are equal.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a