2019 Taiwan APMO Preliminary P3

Art of Problem Solving

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Source: 2019 Taiwan APMO Preliminary

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188 posts

#1 h Aug 23, 2020, 11:27 AM

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Let $\triangle ABC$ be an acute triangle, $H$ is its orthocenter. $\overrightarrow{AH},\overrightarrow{BH},\overrightarrow{CH}$ intersect $\triangle ABC$ 's circumcircle at $A',B',C'$ respectively. Find the range (minimum value and the maximum upper bound) of
$\dfrac{AH}{AA'}+\dfrac{BH}{BB'}+\dfrac{CH}{CC'}$

This post has been edited 1 time. Last edited by hsiangshen, Nov 10, 2020, 2:01 PM

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#2 h Aug 23, 2020, 11:27 AM

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Answer

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#3 h Aug 25, 2020, 11:50 PM

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Bump this post

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#4 h Aug 26, 2020, 12:22 AM

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Why does this need to be bumped? Do you want a solution?

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#5 h Aug 26, 2020, 2:24 AM

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IAmTheHazard wrote:

Why does this need to be bumped? Do you want a solution?

Yes, I can't solve it.

This post has been edited 1 time. Last edited by hsiangshen, Aug 26, 2020, 2:38 AM

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#6 h Aug 26, 2020, 6:32 AM • 1 Y

Y by Mango247

$\delta(H, BC)h_A = bc\cos(B)\cos(C)$ , $h_a = \frac{bc}{2R} = 2R\sin(B)\sin(C)$ , $\delta(H, BC) = 2R\cos(B)\cos(C)$ .

$\frac{AH}{AA'} = \frac{h_A-\delta(H, BC) }{h_A+\delta(H, BC) } = \frac{\cos(A)}{\cos(B-C)}$

$\frac{\cos(A)}{\cos(B-C)} \geq \cos(A)$

Using Law of Cosines on the reflection of C over AH, we get that:

$\frac{\cos(A)}{\cos(B-C)} = \frac{b^2+c^2-a^2}{b^2+c^2-(H_AB-H_AC)^2}$

We also know that $|H_AB-H_AC| = \frac{|b^2-c^2|}{a}$ , so we have:

$\frac{\cos(A)}{\cos(B-C)} = \frac{a^2b^2+a^2c^2-a^4}{a^2b^2+a^2c^2-b^4-c^4+2b^2c^2}$ , cyclic sum is

$(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4)(\frac{1}{a^2b^2+a^2c^2-b^4-c^4+2b^2c^2}+\frac{1}{b^2c^2+c^2a^2-c^4-a^4+2c^2a^2}+\frac{1}{c^2a^2+c^2b^2-a^4-b^4+2a^2b^2}) - 3$

By Cauchy Schwarz, $\text{Sum} \geq \frac{9}{2}-3 = \frac{3}{2}$ .

$\frac{\cos(A)}{\cos(B-C)}=\frac{a^2b^2+a^2c^2-a^4}{a^2b^2+a^2c^2-b^4-c^4+2b^2c^2} < \frac{2a^2b^2+2a^2c^2-2a^4}{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}$ since $\triangle ABC$ is acute, so summing cyclically, $\text{Sum}<2$ , we are done. $\blacksquare$

This post has been edited 2 times. Last edited by Inconsistent, Aug 26, 2020, 6:34 AM
Reason: edit

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#7 h Aug 26, 2020, 7:50 AM

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This has been already posted and I left my solution at the old post.
My solution

This post has been edited 1 time. Last edited by EagleEye, Aug 26, 2020, 7:51 AM

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#8 h Aug 26, 2020, 8:00 AM • 1 Y

Y by Mango247

@above
Thank you two!

This post has been edited 1 time. Last edited by hsiangshen, Aug 26, 2020, 8:00 AM

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