Geometry from Iran TST 2017

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Geometry from Iran TST 2017

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Source: 2017 Iran TST third exam day1 p2

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178 posts

bgn

#1 h Apr 26, 2017, 12:32 PM • 2 Y

Y by plagueis, Adventure10

Let $P$ be a point in the interior of quadrilateral $ABCD$ such that:
$\angle BPC=2\angle BAC \ \ ,\ \ \angle PCA = \angle PAD \ \ ,\ \ \angle PDA=\angle PAC$ Prove that:
$\angle PBD= \left | \angle BCA - \angle PCA \right |$
Proposed by Ali Zamani

This post has been edited 2 times. Last edited by bgn, Apr 27, 2017, 8:19 AM

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bgn

#2 h Apr 27, 2017, 8:21 AM • 1 Y

Y by Adventure10

Edited, I had a typo making the problem wrong.

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4071 posts

#3 h Apr 27, 2017, 11:58 AM • 3 Y

Y by TacH, Adventure10, Mango247

Let's bash!

Inversion with center at $A$ and arbitrary radius. Let $B, C, D, P\mapsto E, F, G, H$ respectively.
Let $\angle{GAH}=a$ and $\angle{HAF}=b$ .
Note that $AFHG$ is a parallelogram and $\angle{EAF}=\angle{BAC}=t, \angle{BPC}=\angle{APC}-\angle{APB}=\angle{AFH}-\angle{AEH}=2t.$ So, $\angle{AEH}=(180^{\circ}-a-b)-2t\implies \angle{EHF}=t$ .
Let $\angle{ABD}=\angle{AGE}=x$ and $\angle{BCA}=\angle{AEF}=y$ .
Note that $\angle{PBD}=|(a+t)-x|$ and $|\angle{BCA}-\angle{PCA}|=|\angle{AEF}-\angle{AHF}|=|y-a|$ .
So, it's enough to prove that $x=y+t$ .

By Ceva's, we've
$\frac{\sin (x)}{\sin (180^{\circ}-x-a-b)}\times \frac{\sin (a+b+t)}{\sin (a+t)}\times \frac{\sin (b+t)}{\sin (a+b+t)}=1$ and $\frac{\sin (y)}{\sin (180^{\circ}-y-a-b-2t)}\times \frac{\sin (t)}{\sin (a)}\times \frac{\sin (b)}{\sin (t)}=1.$
We have
$\begin{align*} & \ \frac{\sin (y+t)}{\sin (180^{\circ}-(y+t)-a-b)}\times \frac{\sin (a+b+t)}{\sin (a+t)}\times \frac{\sin (b+t)}{\sin (a+b+t)}=1\\ & \iff \sin (y+t) \sin (b+t)=\sin (a+b+y+t)\sin (a+t) \\ & \iff \sin (y)\left( \sin (t+b)\cos (t)-\sin (t+a)\cos (a+b+t)\right) =\cos (y)\left( \sin (a+b+t)\sin (t+a)-\sin (t)\sin (t+b)\right)\\ & \iff \sin (y)\left( \frac{(\sin (2t+b)+\sin (b))-(\sin (2t+2a+b)+\sin (-b))}{2} \right) =\cos (y)\left( \frac{(\cos (b)-\cos (b+2a+2t))-(\cos (b)-\cos (2t+b))}{2}\right)\\ & \iff \sin (y)\left( \sin (b)-\frac{\sin (2t+2a+b)-\sin (2t+b)}{2}\right) =\cos (y)\left( -\sin (2t+a+b)\sin (-a)\right)\\ & \iff \sin(y)\left( \sin (b)-\cos (a+b+2t)\sin (a)\right) =\cos (y)\left( \sin (a)\sin (a+b+2t)\right), \end{align*}$ which is true since $\sin (y)\sin (b)=\sin (y+a+b+2t)\sin (a)$ .
So, $\frac{\sin (y+t)}{\sin (180^{\circ}-(y+t)-a-b)} =\frac{\sin (x)}{\sin (180^{\circ}-x-a-b)}.$ This gives $2\sin (y+t)\sin (x+a+b)=2\sin (x)\sin (y+t+a+b)\implies \cos (y+t-x-a-b)=\cos (x-y-t-a-b).$ Hence, we get
$\begin{rcases*} -2\sin(-a-b)\sin(y+t-x)=0 \\ a+b=\angle{DAC}\in (0,\pi ) \\ \begin{rcases*} y+t+a+2b=180^{\circ}\implies y+t\in (0,\pi ) \\ x=\angle{AGE}=\angle{ABC}<180^{\circ}\implies x\in (0,\pi ) \\ \end{rcases*} y+t-x\in (-\pi ,\pi) \\ \end{rcases*} y+t-x=0. \quad \blacksquare$

This post has been edited 4 times. Last edited by ThE-dArK-lOrD, Dec 25, 2018, 3:42 PM

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164 posts

#4 h Apr 27, 2017, 4:54 PM • 3 Y

Y by Adventure10, Mango247, amirhsz

Let $Q$ on $AC$ such that $QA=QB$ and $R\equiv \odot APD \cap PQ$ . It's obvious that $BQPC$ is a cyclic quadrilateral. Now we have $QB^2=QA^2=QR\cdot QP\Rightarrow \overset{\triangle}{QBR}\sim \overset{\triangle}{QPB}\Rightarrow \angle QRB=\angle QBP=\angle ACP=\angle DAP=\angle DRP\Rightarrow B,R,D$ are collinear and so $\angle BCA=\angle BPQ=\angle QBR=\angle QBD=\angle QBP-\angle DBP=\angle ACP-\angle DBP$ and we are done.

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158 posts

#5 h Apr 27, 2017, 7:21 PM • 2 Y

Y by Adventure10, Mango247

Let $J,K$ be the circumcenters of $\triangle ACD , \triangle ABC$ . $O$ lies on $AC$ s.t. $KO \perp AB$ . so $O,P,KC,B$ are concyclic (by simple angle chasing). $BD$ cuts circle $(PBC)$ at $I$ .It's obvious that $A,I,K$ are conllinear. so $\angle IAO=\angle KAI=\angle KCO=\angle AIO$ so $OI=AO=OB$ thus $O$ be the circumcenter of $\triangle ABI$ so $A$ be $C-excenter$ of $\triangle ICB$ . $\angle PBD=\angle PCI=\angle PCA-\angle (ICA=ACB)$ so done.

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2312 posts

#7 h Apr 27, 2017, 10:54 PM • 5 Y

Y by ali.agh, Achillys, enhanced, plagueis, Adventure10

Let $T$ be the point such that $\triangle BPT$ $\stackrel{+}{\sim}$ $\triangle APC$ $\stackrel{+}{\sim}$ $\triangle DPA$ and $S$ be the intersection of $AC,$ $BT.$ Clearly, $S$ lies on $\odot (ABP),$ $(CPT),$ so from $\angle BPC$ $=$ $2\angle BAC$ we get $\angle BAC$ $=$ $\angle BTC$ $\Longrightarrow$ $T$ $\in$ $\odot (ABC),$ hence $\underbrace {\angle PBD = \angle PTA}_{ \because \ \triangle BPT \stackrel{+}{\sim} \triangle DPA}= \angle PTB - \angle ATB = \angle PCA - \angle ACB.$

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PROF65

#8 h Apr 28, 2017, 12:14 PM • 2 Y

Y by ali.agh, Adventure10

unless i'm mistaken the the statement should be: $\angle PBD= \angle BCA + \angle PCA$ with oriented angles .
RH HAS

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Rashed

#11 h Apr 6, 2024, 9:49 AM

Y by

nice problem!
Let K,S be on PB such that KP=PD,SP=PC.
Since ABCS,ABDK are cyclic with easy angle chasing,and power of point satisfies to show PA^2=PK.PS or,PA^2=PD.PC.since P is A_Dumpty point in ADC we are done

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