PQ/AQ =[BC/(AB + AC)]^2, angle and perpendicular bisector related

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PQ/AQ =[BC/(AB + AC)]^2, angle and perpendicular bisector related

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Source: 13th Thailand Mathematical Olympiad 2016 day 1 p1

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#1 h Mar 15, 2020, 11:46 PM

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Let $ABC$ be a triangle with $AB \ne AC$ . Let the angle bisector of $\angle BAC$ intersects $BC$ at $P$ and intersects the perpendicular bisector of segment $BC$ at $Q$ . Prove that $\frac{PQ}{AQ} =\left( \frac{BC}{AB + AC}\right)^2$

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jayme

#2 h Mar 16, 2020, 8:51 AM

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Dear Mathlinkers,
a similar one

https://www.facebook.com/photo.php?fbid=10215351043824082&set=g.1019808738132832&type=1&theater&ifg=1

Sincerely
Jean-Louis

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#3 h Mar 16, 2020, 12:28 PM

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this problem can be done using pop+stewart's theorem+angle bisector theorem, or using the distance formula in barycentric coordinates.
I can post my solution if anyone wants me to do so.

This post has been edited 1 time. Last edited by anyone__42, Mar 16, 2020, 12:32 PM

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#4 h Mar 16, 2020, 3:17 PM

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Let $BC = a, CA = b, AB = c$ . It is known $Q$ is arc midpoint of $BC$ in $(ABC)$ .
$\angle ABP = \angle ABC = \angle AQC = \angle PQC = \angle CQP$
$\angle APB = \angle CPQ$
Thus $\triangle ABP \sim \triangle CQP$ , and $\frac{AP}{BP} = \frac{CP}{QP}$
$PQ = \frac{BP \cdot CP}{AP}$
By Angle Bisector Theorem, $BP = \frac{ac}{b+c}$ and $CP = \frac{ab}{b+c}$ .
So $PQ = \frac{BP \cdot CP}{AP} = \frac{\frac{a^2bc}{(b+c)^2}}{AP}$
$\angle BAQ = \angle PAC$ and $\angle AQB = \angle ACB = \angle ACP$ so $\triangle ABQ \sim \triangle APC$ .
$\frac{AB}{AQ} = \frac{AP}{AC}$
$AQ = \frac{AB \cdot AC}{AP} = \frac{bc}{AP}$
$\frac{PQ}{AQ} = \frac{a^2bc}{bc(b+c)^2} = \frac{a^2}{(b+c)^2} = \left( \frac{BC}{AB + AC}\right)^2$ .

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#5 h Mar 27, 2020, 10:20 PM

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https://i.imgur.com/C413wHk.png

So obviously we have that $Q$ lies on the circle around $\triangle ABC$ .
So I shall start off with a simple angle-chase:
$\angle BAQ = \angle BCQ = \frac{1}{2} \angle A$ Of course the quad $BACQ$ is cyclic, thus Ptolemy's theorem holds, and $AQ$ can be expressed as:
$AQ = BQ.\frac{AC+AB}{BC}$ Now onto some trigonomtry:
From the Angle-Bisector theorem we have that $PC=\frac{AC}{AB} BP$ .
With this in mind,let's apply the sine law on the $\triangle PQC$ and on $\triangle BPQ$ .
Now we get the following results (keep in mind $\angle QPC = \angle C + \frac{1}{2} \angle A$ ):
$PQ=\frac{sin \frac{1}{2} \angle A}{sin(\angle C + \frac{1}{2} \angle A)} BQ$ Now we have the following when we input this into our ratio ( $\frac{PQ}{AQ}$ ):
$\frac{PQ}{AQ} = \frac{sin \frac{1}{2} \angle A}{sin(\angle C + \frac{1}{2} \angle A)}.\frac{BC}{AC+AB}$ Thus we need to show that $\frac{sin \frac{1}{2} \angle A}{sin(\angle C + \frac{1}{2} \angle A)} = \frac{BC}{AC+AB}$
But we applying the sine rule onto the $\triangle ABC$ and representing $\angle B = 180 - \angle A - \angle C$ .This statement get's boiled down to the following:
$1 + cos \angle A = 2cos^2 \frac{1}{2} \angle A$ Which is an identity, thus the statement of the problem is true.....

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Eliti

#6 h Mar 27, 2020, 10:31 PM

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Easy be Stewart's Theorem

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#7 h Mar 27, 2020, 11:14 PM

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Note $Q$ lies on the circumcircle of $\triangle ABC$ . By Ptolemy's theorem on $ABQC$ and since $QB=QC$ , $AQ\cdot BC=(AB+AC)QB\;\Longrightarrow\;\frac{QB}{AQ}=\frac{BC}{AB+AC}.$ $QA\cdot QP=QB^2$ by Shooting lemma (or $\triangle QAB\sim\triangle QBP$ ): we have $\left(\frac{BC}{AB+AC}\right)^2=\frac{QA\cdot QP}{AQ^2}=\frac{PQ}{AQ}.$ $\square$

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nehess

#8 h Jun 10, 2021, 10:08 AM • 2 Y

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By Angle Bisector Theorem: $\frac{PB}{PC} =\frac{AB}{AC} \Rightarrow \frac{PB}{AB} =\frac{PC}{AC} =\frac{BC}{AB+AC}$
We'll prove: $\frac{PQ}{AQ} =(\frac{BC}{AB+AC})^2$
but $(\frac{BC}{AB+AC})^2 =\frac{PB.PC}{AB.AC} =\frac{AP.PQ}{AB.AC}$ since $ABQC$ is cyclic
So it's equivalent to: $AP.AQ=AB.AC$ , which can easy obtain through a $\sqrt{bc}$ inversion

This post has been edited 1 time. Last edited by nehess, Jul 4, 2021, 8:36 AM

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jayme

#9 h Oct 25, 2021, 9:08 AM

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Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/78.%200.%20Relations%20metriques.pdf (Problem 28)

Sincerely
Jean-Louis

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#10 h Oct 26, 2021, 10:17 AM

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jayme wrote:

Dear Mathlinkers,

here (Problem 28)

Sincerely
Jean-Louis

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