LuMaT Prob 03 - Dare for a geo?

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LuMaT Prob 03 - Dare for a geo?

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Source: Indonesian LuMaT 2018 Final Olympiad Day 01 Problem 03

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#1 h Dec 23, 2018, 3:27 PM • 2 Y

Y by Adventure10, Mango247

Given an isosceles acute triangle $ABC$ ( $AB = AC$ ) with circumcircle $\Gamma$ and point $P,Q$ are chosen on segment $BC$ ( $P$ is closer to $B$ than $Q$ ).
Line $AP$ and $AQ$ intersect $\Gamma$ again at $P_0, Q_0$ . Points $P_1, Q_1$ are respectively midpoints of arcs $AP_0, BQ_0$ that contain $B,C$ . Rays $AP_1$ and $AQ_1$ intersects $BC$ at $C_1$ and $B_1$ .
Prove that if the midpoint of $BB_1$ lies on the incircle of $\triangle ACQ$ , then the midpoint of $CC_1$ also lies on the incircle of $\triangle ABP$ .

This post has been edited 2 times. Last edited by IndoMathXdZ, Dec 24, 2018, 1:01 AM

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#2 h Dec 24, 2018, 8:06 AM • 3 Y

Y by Adventure10, Mango247, Mango247

IndoMathXdZ wrote:

Given an isosceles acute triangle $ABC$ ( $AB = AC$ ) with circumcircle $\Gamma$ and point $P,Q$ are chosen on segment $BC$ ( $P$ is closer to $B$ than $Q$ ).
Line $AP$ and $AQ$ intersect $\Gamma$ again at $P_0, Q_0$ . Points $P_1, Q_1$ are respectively midpoints of arcs $AP_0, BQ_0$ that contain $B,C$ . Rays $AP_1$ and $AQ_1$ intersects $BC$ at $C_1$ and $B_1$ .
Prove that if the midpoint of $BB_1$ lies on the incircle of $\triangle ACQ$ , then the midpoint of $CC_1$ also lies on the incircle of $\triangle ABP$ .

As $P$ and $Q$ are independent, for a particular $Q$ satisfying the condition, we need to prove for any point $P$ on segment $BC$ . Therefore we can assume that $Q$ is the midpoint of $BC$ .

Now midpoint of $BB_1$ , say $M$ lies on incircle of $\triangle ACQ$ , so $\frac{CB_1}{2} = MQ = \frac{AQ + QC - CA}{2} \implies CB_1 = AQ + QC - CA.$ Now as $AB_1Q$ is a right isosceles triangle, $QA = QB_1$ so $AC = 2CQ = BC$ or $ABC$ is equilateral.

Now the problem is:

IndoMathXdZ wrote:

Given an equilateral triangle $ABC$ with circumcircle $\Gamma$ and point $P$ on segment $BC$ . Line $AP$ intersect $\Gamma$ again at $P_0$ . Point $P_1$ are midpoints of arcs $AP_0$ that contain $B$ . Ray $AP_1$ intersects $BC$ at $C_1$ . Prove that the midpoint of $CC_1$ lies on the incircle of $\triangle ABP$ .

We need to show $AP + BP - AB + 2PC = CC_1 \Leftrightarrow AP + PC = CC_1 \Leftrightarrow AP = PC_1.$ Notice that $\angle C_1P_1P_0 = \angle C_1P_1B + \angle BP_1P_0 = \angle ABP + \angle BAP = \angle C_1PP_0$ which gives $C_1,P_1,P,P_0$ are concyclic or $\angle PC_1P = \angle P_1P_0P = \angle P_1AP\implies AP + PC_1$ . $\square$

This post has been edited 1 time. Last edited by TheDarkPrince, Dec 28, 2018, 6:23 PM

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#3 h Feb 16, 2021, 5:07 PM

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TheDarkPrince wrote:

IndoMathXdZ wrote:

Given an isosceles acute triangle $ABC$ ( $AB = AC$ ) with circumcircle $\Gamma$ and point $P,Q$ are chosen on segment $BC$ ( $P$ is closer to $B$ than $Q$ ).
Line $AP$ and $AQ$ intersect $\Gamma$ again at $P_0, Q_0$ . Points $P_1, Q_1$ are respectively midpoints of arcs $AP_0, BQ_0$ that contain $B,C$ . Rays $AP_1$ and $AQ_1$ intersects $BC$ at $C_1$ and $B_1$ .
Prove that if the midpoint of $BB_1$ lies on the incircle of $\triangle ACQ$ , then the midpoint of $CC_1$ also lies on the incircle of $\triangle ABP$ .

As $P$ and $Q$ are independent, for a particular $Q$ satisfying the condition, we need to prove for any point $P$ on segment $BC$ . Therefore we can assume that $Q$ is the midpoint of $BC$ .

Now midpoint of $BB_1$ , say $M$ lies on incircle of $\triangle ACQ$ , so $\frac{CB_1}{2} = MQ = \frac{AQ + QC - CA}{2} \implies CB_1 = AQ + QC - CA.$ Now as $AB_1Q$ is a right isosceles triangle, $QA = QB_1$ so $AC = 2CQ = BC$ or $ABC$ is equilateral.

Now the problem is:

IndoMathXdZ wrote:

Given an equilateral triangle $ABC$ with circumcircle $\Gamma$ and point $P$ on segment $BC$ . Line $AP$ intersect $\Gamma$ again at $P_0$ . Point $P_1$ are midpoints of arcs $AP_0$ that contain $B$ . Ray $AP_1$ intersects $BC$ at $C_1$ . Prove that the midpoint of $CC_1$ lies on the incircle of $\triangle ABP$ .

We need to show $AP + BP - AB + 2PC = CC_1 \Leftrightarrow AP + PC = CC_1 \Leftrightarrow AP = PC_1.$ Notice that $\angle C_1P_1P_0 = \angle C_1P_1B + \angle BP_1P_0 = \angle ABP + \angle BAP = \angle C_1PP_0$ which gives $C_1,P_1,P,P_0$ are concyclic or $\angle PC_1P = \angle P_1P_0P = \angle P_1AP\implies AP + PC_1$ . $\square$

There could be a possibility where $Q$ is not the midpoint of $BC$ but still if the midpoint of $BB_{1}$ lies on incircle of $\triangle ACQ$ then midpoint of $CC_{1}$ lies on incircle of $\triangle ABP$ and in this case the triangle might not be equilateral.

The case in which $Q$ is satisfying the given conditions and is also the midpoint of $BC$ , the triangle turns out to be equilateral. So doesn't that mean the above proof would hold only when $Q$ is the midpoint of $BC$ and not other cases

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