Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$.
Problem
Source: Bulgaria MO Regional round 2024, 12.1
Tags: geometry
math_comb01
13.02.2024 16:02
VicKmath7 wrote: Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$. Well Known Lemma: If $ABC$ is a triangle with incircle touching $BC$ at $D$. Let $E$ be on $BC$ and let $I_1,I_2$ denote the incenters of $ABE,ACE$ then $I_1I_2DE$ is cyclic. By the lemma we get $M$ is incircle touch point implying $CA=CB$
ilovemath0402
13.02.2024 18:27
Draw $I_1H, I_2K \perp AB$
Then let $AB = c,BC=a,AC=b,CD=d, AD=x$.
We have $\triangle I_1HM \sim DKI_2 (a-a)$ then $MH.MK = I_1H.I_2K$
We also have $\angle I_1DI_2 = 90^o$ then $\triangle I_1HD \sim DKI_2 (a-a)$ so $DK.DK = I_1H.I_2K$
From two equality we have $MH.MK = DH.DK (1)$
But
$MH = AM - AH = \dfrac{c}{2} - \dfrac{b+x-d}{2} = \dfrac{c-b-x+d}{2}$
$MK = MB-BK = \dfrac{c}{2} - \dfrac{a+c-x-d}{2} = \dfrac{x+d-a}{2}$
$DH = \dfrac{x+d-b}{2}$
$DK = \dfrac{d+c-x-a}{2}$
So (1) become $(c-b-x+d)(x+d-a)=(x+d-b)(c+d-x-a)$
Let $c+d-x=u,x+d=v$. The above become $(u-b)(v-a)=(v-b)(u-a) \Leftrightarrow (a-b)(u-v)=0$.
If $u=v$ then $c=2x \Leftrightarrow AM = AD$ (we can't have this because $D \not \equiv M$. So $a=b$. And we have $CA=CB$
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