Problem 3 of First round

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Problem 3 of First round

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Source: V International Festival of Young Mathematicians Sozopol 2014, Theme for 10-12 grade

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Pinko

#1 h Oct 9, 2019, 7:20 AM • 1 Y

Y by Adventure10

In an acute $\Delta ABC$ , $AH_a$ and $BH_b$ are altitudes and $M$ is the middle point of $AB$ . The circumscribed circles of $\Delta AMH_a$ and $\Delta BMH_b$ intersect for a second time in $P$ . Prove that point $P$ lies on the circumscribed circle of $\Delta ABC$ .

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#2 h Oct 9, 2019, 9:31 PM • 2 Y

Y by Pinko, Adventure10

Solution

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LKira

#3 h Oct 9, 2019, 10:01 PM • 1 Y

Y by Adventure10

Ricochet wrote:

Solution

I think your solution is correct
Btw, we can just use the inversion center $H$ swap $A$ and $H_a$

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#4 h Oct 9, 2019, 10:08 PM • 1 Y

Y by Adventure10

Pinko wrote:

In an acute $\Delta ABC$ , $AH_a$ and $BH_b$ are altitudes and $M$ is the middle point of $AB$ . The circumscribed circles of $\Delta AMH_a$ and $\Delta BMH_b$ intersect for a second time in $P$ . Prove that point $P$ lies on the circumscribed circle of $\Delta ABC$ .

First of all note that power of $H$ wrt both the circles is same hence $H, M, P$ collinear. Now since $MH_a=MH_b=MA=MB$ hence $MP$ bisects $\angle APH_a$ and $\angle BPH_b$ .Hence $\dfrac{PA}{PH_a}=\dfrac{PB}{PH_b} \implies \dfrac{PA}{PB}=\dfrac{PH_a}{PH_b}$ hence since $ABH_aH_b$ is cyclic we obtain by some trivial calculation that $P$ is center of spiral similarity taking $H_aH_b \to BA$ hence $P\in \odot{ABC}$ as desired.

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#5 h Oct 10, 2019, 12:04 AM • 3 Y

Y by Pinko, Adventure10, Mango247

We let $D$ be the reflection of the orthocenter $H$ across $H_a$ , $E$ be the reflection of $H$ across $H_b$ , and $F$ be the reflection of $H$ across $M$ . Note that $D, E, F$ are on $(ABC)$ .
Claim: $P$ lies on line $MH$ .
Proof: $MP$ is the radical axis of $(AMH_a)$ and $(BMH_b)$ .
So we just have to prove $H$ lies on the radical axis, i.e. has equal power wrt both circles.
$Pow_{(AMH_a)}H = HA*HH_a$
$Pow_{(BMH_b)}H = HB*HH_b$
But since $\angle AH_aB = \angle AH_bB = 90^{\circ}$ , $ABH_aH_b$ is cyclic.
So $HA*HH_a = HB*HH_b$ , so $H$ lies on the radical axis and the claim is proved.

$HP*HF = 2HM*HP = 2HA*HH_a =HA*HD$
So $AFDP$ is cyclic, but $(AFD)$ is $(ABC)$ , so $P$ lies on $(ABC)$ .

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