2012 Japan Mathematical Olympiad Finals Problem 4

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2012 Japan Mathematical Olympiad Finals Problem 4

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Feb 11, 2012, 5:32 PM • 2 Y

Y by Adventure10, Mango247

Given two triangles $PAB$ and $PCD$ such that $PA=PB,\ PC=PD,\ P,\ A,\ C$ and $B,\ P,\ D$ are collinear in this order respectively.
The circle $S_1$ passing through $A,\ C$ intersects with the circle $S_2$ passing through $B,\ D$ at distinct points $X,\ Y$ .
Prove that the circumcenter of the triangle $PXY$ is the midpoint of the centers of $S_1,\ S_2$ .

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#2 h Feb 11, 2012, 6:47 PM • 2 Y

Y by Adventure10, Mango247

Here is a general problem

Let $ABC$ be triangle. $M$ is a point on side $AB$ . $N$ is a point on $AC$ such that $MN$ is parallel to $A$ -symmedian of triangle $ABC$ . The circle $(O_1)$ passing through $M,B$ intersects circle $(O_2)$ passing through $N,C$ at t distinct points $X, Y$ . Prove that circumcenter of triangle $AXY$ is midpoint of $O_1O_2$ .

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#3 h Feb 11, 2012, 10:23 PM • 3 Y

Y by WakeUp, Adventure10, Mango247

buratinogigle wrote:

Let $ABC$ be triangle. $M$ is a point on side $AB.$ $N$ is a point on $AC$ such that $MN$ is parallel to $A$ -symmedian of triangle $ABC.$ The circle $(O_1)$ passing through $M,B$ intersects circle $(O_2)$ passing through $N,C$ at t distinct points $X, Y.$ Prove that circumcenter of triangle $AXY$ is midpoint of $O_1O_2.$

Let $K$ denote the foot of the A-symmedian and $D \equiv MN \cap BC.$ Then

$\frac{\overline{AN}}{\overline{AC}}=\frac{\overline{KD}}{\overline{KC}} \ , \ \frac{\overline{AM}}{\overline{AB}}=\frac{\overline{KD}}{\overline{KB}} \Longrightarrow \frac{\overline{AN} \cdot \overline{AC}}{\overline{AM} \cdot \overline{AB}}=\frac{\overline{KD}}{\overline{KC}} \cdot \frac{\overline{KB}}{\overline{KD}} \cdot \frac{\overline{AC}^2}{\overline{AB}^2}=-1$

This means that the ratio of the powers of $A$ WRT $(O_1)$ and $(O_2)$ equals $-1$ $\Longrightarrow$ $A$ is on the circle coaxal with $(O_1),$ $(O_2)$ whose center $O$ divides $\overline{O_1O_2}$ in the ratio $\overline{OO_1}:\overline{OO_2}=-1,$ i.e. circumcenter $O$ of $\triangle AXY$ is midpoint of $\overline {O_1O_2}.$

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#4 h Feb 13, 2012, 3:52 PM • 7 Y

Y by WakeUp, mathematiculperson, Adventure10, Mango247, and 3 other users

Let $O_1$ , $O_2$ be the circumcenters of $S_1$ and $S_2$ , respectively. Let $M$ be the midpoint of $O_1 O_2$ and let $MO_1 = MO_2 = m$ . Let $r_1$ and $r_2$ be the radii of $S_1$ and $S_2$ , respectively. By power of a point, it follows that $PA \cdot PC = PO_1^2 - r_1^2$ and $PB \cdot PD = r_2^2 - PO_2^2$ and hence that $PO_1^2 + PO_2^2 = r_1^2 + r_2^2$ . Also note that $XO_1^2 + XO_2^2 = YO_1^2 + YO_2^2 = r_1^2 + r_2^2$ . By Stewart's theorem, it now follows that $2PM^2 + 2m^2 = PO_1^2 + PO_2^2 = r_1^2 + r_2^2 = 2XM^2 + 2m^2 = 2YM^2 + 2m^2$ . Hence $XM = YM = PM$ and $M$ is the circumcenter of $\triangle{PXY}$ .

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#5 h Feb 28, 2012, 3:25 PM • 2 Y

Y by Adventure10, Mango247

Let $K$ be the mipoint of segment $S_{1}S_{2}$ ,then by $PB*PD=PA*PC$ ,we get $R_{1}^{2}-O_{1}P^{2}=O_{2}P^{2}-R_{2}P^{2}$ ,
then we have $PK^{2}=R_{1}^{2}+R_{2}^{2}-\frac{1}{2}OO_{2}^{2}=XK^{2}=YK^{2}$

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#6 h Sep 13, 2022, 11:39 PM • 2 Y

Y by teomihai, ike.chen

The problem is essentially one key lemma.

Lemma. The locus of points $P$ such that the sum of powers of $P$ to two fixed circles $\omega_1$ and $\omega_2$ is centered at the midpoint of the segment connecting the centers of these two circles.

Proof. Let $M$ be the midpoint and $O_1, O_2$ be the two centers. By the median length formula, $PM = \frac{\sqrt{2PO_1^2+2PO_2^2-O_1O_2^2}}2$ is constant, yay.

Now take $S_1$ and $S_2$ to be the circles and notice that the sum of powers of $P$ to these two circles is equal to $0$ by the lengths condition. Thus, $P, X, Y$ all lie on a circle centered at the midpoint of the centers by the lemma, as desired.

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#7 h Sep 14, 2022, 12:05 AM

Y by

HamstPan38825 wrote:

The problem is essentially one key lemma.

Lemma. The locus of points $P$ such that the sum of powers of $P$ to two fixed circles $\omega_1$ and $\omega_2$ is centered at the midpoint of the segment connecting the centers of these two circles.

Proof. Let $M$ be the midpoint and $O_1, O_2$ be the two centers. By the median length formula, $PM = \frac{\sqrt{2PO_1^2+2PO_2^2-O_1O_2^2}}2$ is constant, yay.

Now take $S_1$ and $S_2$ to be the circles and notice that the sum of powers of $P$ to these two circles is equal to $0$ by the lengths condition. Thus, $P, X, Y$ all lie on a circle centered at the midpoint of the centers by the lemma, as desired.

Splendid solution!

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#8 h Jun 9, 2023, 9:47 PM

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My solution is mostly equivalent to post #3, but I’d like to add that there exists a solution to ELMO 2017/2 whose set-up is essentially the converse of this result.

This post has been edited 6 times. Last edited by ike.chen, Jun 30, 2023, 6:25 AM

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#10 h Jul 15, 2023, 8:06 PM

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Nice problem for using pascals midpoint theorem!

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