Bosnia and Herzegovina 2018 TST Day 1 Problem 1

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Bosnia and Herzegovina 2018 TST Day 1 Problem 1

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Bosnia and Herzegovina 2018 TST

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

741 posts

#1 h Sep 15, 2018, 10:52 PM • 1 Y

Y by Adventure10

In acute triangle $ABC$ $(AB < AC)$ let $D$ , $E$ and $F$ be foots of perpedicular from $A$ , $B$ and $C$ to $BC$ , $CA$ and $AB$ , respectively. Let $P$ and $Q$ be points on line $EF$ such that $DP \perp EF$ and $BQ=CQ$ . Prove that $\angle ADP = \angle PBQ$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2 posts

jerker

#2 h Oct 4, 2018, 12:46 AM • 2 Y

Y by Adventure10, Mango247

Using the property of complete quadrilateral we have BQPC concyclic.Because ADP=C-B so only prove QBP=C-B FPB=x then we can easily have PBC=x+B-C HBC=x which is exactly what we are gonna prove.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

472 posts

Ari004

#3 h Oct 16, 2018, 1:13 PM • 2 Y

Y by Adventure10, Mango247

Let $EF$ meet $BC$ at $T$ .Since, $(C,B;D,T)=-1$ and $DP \perp EF,DP$ must be the internal bisector of $\angle BPC$ .As $Q$ lies on the perpendicular bisector of $BC,Q$ must be the midpoint of arc $BPC$ .Hence, $B,P,Q,C$ are concyclic.By angle chase, $\angle ADP=\angle B-\angle C$ .Let $\angle BPF=\alpha$ .Then, $\angle BCQ=\angle CBQ=\alpha,\angle BPD=90^{\circ}-\alpha,\angle BDP=90^{\circ}+\angle C-\angle B$ .All these imply $\angle PBQ=\angle B-\angle C$ and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1218 posts

#4 h Oct 17, 2018, 11:15 AM • 2 Y

Y by Adventure10, Mango247

My solution: WLOG assume that $AB \leq AC$ . Let $M$ be the midpoint of $BC$ , and let $EF \cap BC=X$ . Note that $PDMQ$ is cyclic. Using the converse of the Lemma here (which can be proved in a similar way as the proof of the Lemma itself in this link), we get $XB \cdot XC=XD \cdot XM=XP \cdot XQ$ , giving that $BPQC$ is cyclic. As $QB=QC$ , this gives that $PQ$ is the external angle bisector of $\angle BPC$ .

Now, Let $\ell$ be the line through $Q$ perpendicular to $EF$ . Then as $\ell \parallel DP$ and $QM \parallel DA$ , we get that $\angle ADP=\angle (\ell,QM)=90^{\circ}-\angle PQM=90^{\circ}-\angle BQM-\angle PQB=\angle CBQ-\angle PQB$ And, $\angle CBQ=\angle CPQ=\angle FPB$ . Thus $\angle PBQ=\angle FPB-\angle PQB=\angle ADP \text{ } \blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2125 posts

#5 h Jul 8, 2019, 9:38 AM • 3 Y

Y by Pluto1708, Adventure10, Mango247

From Harmonic Bundle argument, obviously $PD$ is angle bisector of $\angle BPC$ . Let $\odot (BPC)$ $\cap$ $EF$ $=$ $Q'$ , then, $Q'$ is the midpoint of arc $BPC$ $\implies$ $Q' \equiv Q$ . Let $PD$ $\cap$ $\odot (BPC)$ $=$ $D'$ $\implies$ $\angle ADP$ $=$ $\angle PD'Q$ $=$ $\angle PBQ$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1915 posts

#6 h Jan 14, 2020, 12:23 PM • 3 Y

Y by GeoMetrix, strawberry_circle, Adventure10

Maybe same as others

FBH 2018 TST Day 1 P1 wrote:

In acute triangle $ABC$ $(AB < AC)$ let $D$ , $E$ and $F$ be foots of perpedicular from $A$ , $B$ and $C$ to $BC$ , $CA$ and $AB$ , respectively. Let $P$ and $Q$ be points on line $EF$ such that $DP \perp EF$ and $BQ=CQ$ . Prove that $\angle ADP = \angle PBQ$

Let $M$ be the midpoint of $BC$ . So $QM\perp BC$ . Hence, $D,P,Q,M$ are concyclic. Let $EF\cap BC=X_A$ . Then as $(X_AD;BC)=-1$ . Hence, by Maclaurin's Theorem we get that $X_AB\cdot X_AC=X_AD\cdot X_AM=X_AP\cdot X_AQ\implies B,P,Q,C\text{ are concyclic.}$ Let $QM\cap PD=T$ . $PD$ is the internal bisector of $\angle BPC$ as $(X_AD;BC)=-1$ and $\angle DPX_A=90^\circ$ . So, $P,B,T,C$ are also conyclic. Hence, $P,Q,C,T,B$ are all concyclic. So, $\angle ADP=\angle QX_AC=\angle PTQ=\angle PCQ=\angle PBQ.\blacksquare$

This post has been edited 3 times. Last edited by amar_04, Jan 14, 2020, 12:25 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41 posts

#7 h Aug 20, 2020, 1:19 PM

Y by

Firstly, let's denote $M$ as midpoint of $BC$ and let $\ell$ be perpendicular bisector of $BC$ . Then we know that $Q = \ell \cap EF$ . So from there $\angle QMD = \angle QPD = 90^\circ$ hence $D, P, M, Q$ are concyclic. Let $G = EF \cap BC$ . Now lets denote circumcircle of $DPMQ$ as $\omega_1$ . Now we have
$Pow_{\omega_1}(G) = GP \cdot GQ = GD \cdot GM$ We want to prove that $B, P, C, Q$ are concyclic. To do that we are left to prove
$GD \cdot GM = GB \cdot GC$ Which follows from Manclaurin's theorem because $(GD;BC) = - 1$ .
From this we conclude that $BPCQ$ is cyclic. Easily we obtain $\angle APD = \angle B - \angle C$ . Let $BPF = x$ . Then $angle BCQ = \angle CBQ = x$ , and so
$\angle BPD = 90^{\circ} - x, \angle BDP = 90^circ - \angle B + \angle C$ This implies $\angle PBQ = \angle B - \angle C$ thus we are done.

This post has been edited 1 time. Last edited by AleksaS, Aug 20, 2020, 1:20 PM
Reason: Latex

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a