Equal Distances in an Isosceles Setting

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Equal Distances in an Isosceles Setting

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Source: IDMC 2025 P4

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#1 h Apr 20, 2025, 5:05 AM • 1 Y

Y by sami1618

Let $ABC$ be an isosceles triangle with $AB=AC$ . The circle $\omega_1$ , passing through $B$ and $C$ , intersects segment $AB$ at $K\neq B$ . The circle $\omega_2$ is tangent to $BC$ at $B$ and passes through $K$ . Let $M$ and $N$ be the midpoints of segments $AB$ and $AC$ , respectively. The line $MN$ intersects $\omega_1$ and $\omega_2$ at points $P$ and $Q$ , respectively, where $P$ and $Q$ are the intersections closer to $M$ . Prove that $MP=MQ$ .

Proposed by Hooman Fattahi

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#2 h Apr 20, 2025, 6:14 AM • 1 Y

Y by sami1618

Not really a fan of this solution but.. whatever I guess :-D

:-D

Let $MN$ intersect $\omega_1$ and $\omega_2$ again at $X, Y$ such that $M$ lies between $P$ and $X$ at the same time, between $Q$ and $Y$ . Denote $E, F$ the midpoints of $PX$ and $QY$ .
As per the definition of $\omega_2$ , We have that $BE$ passes through the center of $\omega_2$ , which means $\triangle BPX$ is isoceles. By symmetry $\triangle AQY$ is also isoceles. We have that $AEBF$ is a parallelogram, thus $\overline{ME} + \overline{MF} = 0$ . Or in other words:
$\overline{MP} + \overline{MQ} + \overline{MX} + \overline{MY} = 0$ But then, by power of a point: $\overline{MP}\cdot\overline{MX} = \overline{MK}\cdot\overline{MB} = \overline{MQ}\cdot\overline{MY}$
Let $\overline{MP} = a, \overline{MQ} = b, \overline{MX} = c, \overline{MY} = d$ .
At this point: $a + b + c + d = 0$ and $ac = bd$ . That being said, $bd + (b + c + d)\cdot c = 0 \leftrightarrow (c + b)(c + d) = 0$ .
The case $c = -b$ is obviously ruled out as $P, M, X$ are collinear in that very same order.
Hence, $c = -d$ , we'll have $a = -b$ or $MP = MQ$ , thus proving the problem.

This post has been edited 2 times. Last edited by Retemoeg, Apr 20, 2025, 6:20 AM

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Mahdi_Mashayekhi

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#3 h Apr 20, 2025, 8:28 AM • 1 Y

Y by sami1618

Let $PQ$ meet $\omega_1$ and $\omega_2$ again at $X$ and $Y$ . Let $MX = \alpha, MY = \beta, MP = \theta, MQ = \lambda$ Now let $f(X) = XP^2-XQ^2$ . By linearity of PoP we have $f(M)=\frac{\theta (\theta+\lambda)^2-\lambda (\theta+\lambda)^2}{\theta+\lambda}=(\theta+\lambda)(\theta-\lambda)$ and $f(M)=\frac{\alpha f(X) + \beta f(Y)}{\alpha+\beta}=\frac{\alpha (\theta+\lambda)(\theta-\lambda+2\alpha)-\beta (\theta+\lambda)(\lambda-\theta+2\beta))}{\alpha+\beta}$ so we have $\frac{\alpha(\theta-\lambda+2\alpha)-\beta(\lambda-\theta+2\beta)}{\alpha+\beta}=(\theta-\lambda)$ which implies $\alpha^2-\beta^2=0 \implies \alpha=\beta$ . Now note that $MP.MX=MK.MB=MQ.MY$ so $\alpha\theta=\beta\lambda$ so now we have $\theta=\lambda$ as wanted.

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#4 h Apr 20, 2025, 9:45 PM • 1 Y

Y by Retemoeg

Let line $MN$ cut $\omega_1$ and $\omega_2$ for the second time at $P'$ and $Q'$ , respectively. Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively. Finally, let $M_p$ be the midpoint of $PP'$ and $M_q$ be the midpoint $QQ'$ .
$[asy] import geometry; size(10cm); pair A=dir(90); pair B=dir(220); pair C=dir(320); pair K=B+.6(A-B); pair O_1=circumcenter(B,K,C); circle w_1=circle(K,B,C); pair M=.5(A+B); pair N=.5(A+C); point[] T=intersectionpoints(w_1,line(N,M)); pair P=T[0]; pair Pp=T[1]; line b=line(.5(B+K),.5(B+K)+rotate(90)*(B-K)); line l=line(B,B+rotate(90)*(C-B)); pair O_2=intersectionpoint(b,l); circle w_2=circle(B,K,O_2+rotate(90)*(K-O_2)); point[] X=intersectionpoints(w_2,line(M,N)); pair Q=X[1]; pair Qp=X[0]; pair M_p=.5(P+Pp); pair M_q=.5(Q+Qp); fill(A--B--C--cycle, palered); draw(A--B--C--cycle); draw(w_1,deepblue); draw(w_2,deepgreen); draw(Pp--Qp); draw(B--M_q, dashed); draw(O_1--A, dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$K$",K,2*dir(90)); dot("$O_1$",O_1,dir(270)); dot("$M$",M,dir(120)); dot("$N$",N,dir(60)); dot("$P$",P,dir(120)); dot("$P'$",Pp,dir(60)); dot("$O_2$",O_2,dir(180)); dot("$Q$",Q,dir(60)); dot("$Q'$",Qp,dir(120)); dot("$M_p$",M_p,dir(60)); dot("$M_q$",M_q,dir(120)); [/asy]$
Since $O_1M_p\perp MN\parallel BC$ , it follows that $M_p$ is also the midpoint of $MN$ . Similarly, $M_q$ is the foot of $B$ onto $MN$ , so $|M_pM|=|M_qM|$ . Then $|M_pP|^2-|M_pM|^2=|MP|\cdot|MP'|=-\mathbb{P}(M,\omega_1)=|MK|\cdot |MB|=$ $-\mathbb{P}(M,\omega_2)=|MQ|\cdot|MQ'|=|M_qQ|^2-|M_qM|^2,$ so $|M_pP|=|M_qQ|$ . The result now easily follows (if $K$ does not lie on segment $AM$ then the signs would need to be reversed).

This post has been edited 6 times. Last edited by sami1618, Apr 20, 2025, 10:06 PM

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