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Source: Pan-American Girls’ Mathematical Olympiad 2022, Problem 3

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122 posts

#1 h Oct 27, 2022, 8:11 PM • 4 Y

Y by LLL2019, Mango247, lian_the_noob12, Rounak_iitr

Let $ABC$ be an acute triangle with $AB< AC$ . Denote by $P$ and $Q$ points on the segment $BC$ such that $\angle BAP = \angle CAQ < \frac{\angle BAC}{2}$ . $B_1$ is a point on segment $AC$ . $BB_1$ intersects $AP$ and $AQ$ at $P_1$ and $Q_1$ , respectively. The angle bisectors of $\angle BAC$ and $\angle CBB_1$ intersect at $M$ . If $PQ_1\perp AC$ and $QP_1\perp AB$ , prove that $AQ_1MPB$ is cyclic.

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6870 posts

#2 h Oct 27, 2022, 11:13 PM • 4 Y

Y by HamstPan38825, LLL2019, teomihai, crazyeyemoody907

We first delete several useless points.

We delete point $B_1$ because it does nothing.
Proving $AQ_1PB$ was cyclic would solve the problem as then $M$ would be simply be the midpoint of its arc $Q_1P$ . Thus, we can ignore $M$ as well.
Since $\measuredangle Q_1PP_1 = 90^{\circ} - \measuredangle PAC = 90^{\circ} - \measuredangle BAQ = \measuredangle Q_1QP_1$ so $PP_1Q_1Q$ is cyclic, and we can delete point $C$ too.

So after cutting away all the smoke and mirrors, the problem says: let $PP_1Q_1Q$ be cyclic with $A = \overline{P_1P} \cap \overline{Q_1Q}$ and $B = \overline{PQ} \cap \overline{P_1Q_1}$ . Given that $\overline{P_1Q} \perp \overline{AB}$ , prove $AQ_1PB$ is cyclic.

$[asy] size(11cm); pair A = dir(110); pair B = dir(210); pair Q = dir(330); pair P = foot(A, B, Q); pair P_1 = orthocenter(A, B, Q); pair Q_1 = foot(B, A, Q); pair X = foot(Q, A, B); pair Y = foot(A, P, Q_1); pair C = extension(A, Y, P, Q); pair B_1 = extension(B, Q_1, A, C); draw(Q--X); draw(A--P--Q_1); draw(B--Q_1); filldraw(A--B--Q--cycle, invisible, blue); filldraw(circumcircle(P, P_1, Q), invisible, red); draw(Q_1--B_1, dotted+grey); draw(Q_1--Y, dotted+grey); draw(A--C--Q, dotted+grey); draw(circumcircle(A, P, B), grey); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); dot("$P_1$", P_1, 1.4*dir(120)); dot("$Q_1$", Q_1, dir(Q_1)); dot(X); dot(Y); dot("$C$", C, dir(C)); dot("$B_1$", B_1, dir(B_1)); /* TSQ Source: !size(11cm); A = dir 110 B = dir 210 Q = dir 330 P = foot A B Q P_1 = orthocenter A B Q 1.4R120 Q_1 = foot B A Q X .= foot Q A B Y .= foot A P Q_1 C = extension A Y P Q B_1 = extension B Q_1 A C Q--X A--P--Q_1 B--Q_1 A--B--Q--cycle 0.1 yellow / blue circumcircle P P_1 Q 0.1 orange / red Q_1--B_1 dotted grey Q_1--Y dotted grey A--C--Q dotted grey circumcircle A P B grey */ [/asy]$
Well, Brokard's theorem says $\overline{AB}$ is the polar of $\overline{PQ_1} \cap \overline{P_1Q}$ , so that forces $(P_1PQQ_1)$ to have diameter $\overline{P_1Q}$ . It follows $P_1$ is exactly the orthocenter of $\triangle ABQ$ , and $AQ_1PB$ is cyclic.

This post has been edited 7 times. Last edited by v_Enhance, Oct 30, 2022, 6:34 PM

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834 posts

#3 h Oct 30, 2022, 3:50 AM • 3 Y

Y by teomihai, Mango247, Mango247

The problem seems to be having weird wording to cover up the crux beneath it, but this makes me not like the problem so much.

We will now reconstruct the problem:
$ABC$ is an acute triangle with $AB<AC.$ $P,Q$ lie on $BC$ such that $AP$ and $AQ$ are isogonal. $P_1$ and $Q_1$ lie on $AP$ and $AQ$ respectively such that $P_1Q$ is perpendicular to $AB$ and $Q_1P$ is perpendicular to $AC.$ If $BP_1Q_1$ is collinear, prove that $AQ_1PB$ is cyclic.

Note that this is equivalent, as $M$ lying on the circle would hold by it being the midpoint of the arc $PQ_1.$

Now we prove that the condition is also equivalent to $AP$ being perpendicular to $BC.$

$PP_1Q_1Q$ is cyclic. This is since $\measuredangle P_1Q_1Q=\measuredangle BQ_1A=\measuredangle QPP_1.$ (in fact this is true without the collinear condition, you can try to prove this)

Now angle chase. $\measuredangle P_1QP=\measuredangle P_1QP=\measuredangle BAP=\measuredangle QPP_1-\measuredangle CBA.$ However, $\measuredangle P_1QP=90^{\circ}-\measuredangle CBA,$ thus we must have $P$ be the foot of altitude.

Now, we want to prove $\angle AQ_1B=90^{\circ},$ but this is easy because $PQQ_1P_1$ is cyclic.

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1080 posts

#4 h Nov 30, 2022, 6:39 AM • 2 Y

Y by centslordm, Mango247

Lemma: Let $H$ be on the $A$ -altitude of $\triangle ABC$ and let $E=\overline{BH}\cap\overline{AC}$ and $F=\overline{CH}\cap\overline{AB}.$ If $A,E,F,H$ are concyclic, then $H$ is the orthocenter of $\triangle ABC.$
Proof. Let $X=\overline{AH}\cap\overline{EF}$ and by Brokard, the center of $(AEHF)$ is the orthocenter of $\triangle XBC.$ Notice that the aforementioned orthocenter lies on $\overline{XH}$ as $\overline{XH}\perp\overline{BC}.$ Hence, $\overline{AH}$ is a diameter of $(AEHF)$ and $\angle AEH=90.$ $\blacksquare$

Note $\measuredangle PP_1Q_1=90-\measuredangle BAP_1=90-\measuredangle Q_1AC=\measuredangle PQ_1Q$ so $PQQ_1P_1$ is cyclic and $P_1$ is the orthocenter of $\triangle ABQ$ by our lemma. Therefore, $ABPQ_1$ is cyclic and $\angle MBQ_1=\tfrac{1}{2}\angle PBQ_1=\tfrac{1}{2}\angle PAQ_1=\angle PAM$ so $AMPB$ is cyclic. $\square$

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CoolCarsOnTheRun

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CoolCarsOnTheRun

#5 h Dec 29, 2022, 11:21 PM • 1 Y

Y by centslordm

Let line $P_1Q$ meet line $AB$ at point $X$ and line $PQ_1$ meet line $AC$ at point $Y$ . As $\angle{XAP_1}=\angle{YAQ_1}$ , we know $\triangle{XAP_1}\sim\triangle{YAQ_1}$ . Thus, $\angle{PP_1Q}=\angle{XP_1A}=\angle{YQ_1A}=\angle{QQ_1P}$ , so quadrilateral $PP_1Q_1Q$ is cyclic. Therefore, $Z$ is the pole of line $AB$ by Brokard's Theorem on $PP_1QQ_1$ . But as $XP_1\perp AB$ , we know that the center of $(PP_1Q_1Q)$ lies on line $XP_1$ , so $P_1Q$ is a diameter of $(PP_1Q_1Q)$ . Hence, $\angle{P_1PQ}=\angle{P_1Q_1Q}=90^\circ$ , so $\angle{BPA}=\angle{BQ_1A}=90^\circ$ , and thus quadrilateral $ABPQ_1$ is cyclic.

Finally, $M$ clearly lies on $(ABPQ_1)$ as the midpoint of arc $PQ_1$ , completing the proof.

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4175 posts

#6 h Jan 21, 2023, 10:31 PM • 1 Y

Y by Mango247

Let $PQ_1$ intersect $AC$ at $S$ and $QP_1$ intersect $AB$ at $R$ .

Claim: $PQ_1QP$ is cyclic. We have, $\angle BAP=\angle CAQ,$ so $\angle RP_1A=\angle SQ_1A$ . Consequently, $\angle PP_1Q=\angle PQ_1Q,$ so $PQ_1QP$ is cyclic.

Let $O$ be the circumcenter of $PQ_1QP$ , and let $P_1Q$ intersect $PQ_1$ at $T.$ By Brokard, $O$ is the orthocenter of $\triangle BAT,$ so $O$ lies on the perpendicular from $T$ to line $AB$ , which is also $P_1Q$ , so $O$ lies on $P_1Q$ . However, since $OP_1=OQ$ , this means that $O$ is the midpoint of $P_1Q$ , so we have $\angle P_1PQ=\angle P_1Q_1Q=90,$ so $AQ_1PB$ is cyclic since $\angle AQ_1B=\angle APB=90.$

From $AQ_1PB$ cyclic, we also have $\angle PBQ_1=\angle PAQ_1.$ However, we also have $\angle MAP=\angle MBP=\frac{1}{2}\angle PAQ_1=\frac{1}{2}\angle PBQ_1,$ so $ABPM$ is cyclic. Since $AQ_1PB$ is cyclic and $ABPM$ is cyclic, we are done.

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465 posts

#7 h May 25, 2023, 6:26 PM • 1 Y

Y by HoripodoKrishno

This is really good :surrender:

:surrender:

Also wow, it seems that I have literally forgotten the `a' of angel chessing :surrender:

:surrender:

https://i.imgur.com/I5YpJfZ.png

Firstly, let $E=PQ_1\cap AC$ and $F=P_1Q\cap AB$ . Now note that $\measuredangle P_1PQ_1=\measuredangle APE=\measuredangle PAE+\measuredangle AEP=\measuredangle BAQ+90^\circ=\measuredangle FAQ+\measuredangle QFA=\measuredangle P_1QQ_1\implies PQQ_1P_1$ is cyclic.

Now applying Brokard theorem on $\odot(PQQ_1P_1)$ gives that as $PQ\cap P_1Q_1$ lies on $P_1Q$ , so the center also must lie on $P_1Q$ as the line $AB$ then becomes perpendicular to line through $O$ (the center) and $P_1Q\cap PQ_1$ and so this gives that $\measuredangle P_1PQ=\measuredangle P_1Q_1Q=90^\circ$ which further gives $\measuredangle APB=\measuredangle P_1PQ=90^\circ=\measuredangle BQ_1A\implies ABPQ_1$ is cyclic.

Now to finish, note that $M$ is now just the intersection of the angle bisectors of $\angle PAQ$ and $\angle QBQ_1$ and we have that $\measuredangle AMB=90^\circ$ from the following lemma from here which finishes.

Lemma wrote:

Let $ABCD$ be a cyclic quadrilateral. Let $U=AB\cap CD$ and $V=AD\cap BC$ . $\ell_1$ and $\ell_2$ denote the angle bisectors of $\angle AUD$ and $\angle AVB$ . Then $\ell_1\perp\ell_2$ .

Proof follows from direct angle chasing.

https://i.imgur.com/t9NVCAC.png

This post has been edited 1 time. Last edited by kamatadu, May 25, 2023, 6:27 PM

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#9 h Aug 10, 2023, 5:36 PM

Y by

Great problem! Let D,E,F be the intersection of P_1Q with AB, PQ_1 with AC, and P_1Q with PQ_1, respectively. First notice QP_1P=AP_1D=90-DAP_1=90-Q_1AE=AQ_1E=QQ_1P, hence PP_1Q_1Q is cyclic. Next, let O be the circumcenter of that quadrilateral. From Brocard's theorem, O is the orthocenter of P_1AB (from complete quadrilateral AP_1BPCQ_1), hence OP_1 perp. AB. Since DP_1 also perp. AB, O lies on the line DP_1Q; in particular, since O is the circumcenter it's the midpoint of P_1Q, implying that P_1Q is a diameter and hence P_1Q_1Q=P_1PQ=90 degrees. This implies a lot of things:

AQ_1B is 90 degrees, whence Q_1 lies on the circle with diameter AB, analogously so does P. Now note that AMB is 90 degrees, whence M lies on this circle too, and we are done!
note that B_1 was not used

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1500 posts

#10 h Aug 30, 2023, 2:33 AM

Y by

Who let them cook.

Note that if we show that $AQ_1PB$ is cyclic, the fact that $M$ lies on the circle follows immediately.

Claim: Quadrilateral $PP_1QQ_1$ is cyclic.
Proof. Follows since $\measuredangle PP_1Q = \measuredangle AP_1F = \measuredangle P_1AF + 90^\circ$ and similarily $\measuredangle PQ_1Q = GAQ_1 + 90^\circ$ . $\blacksquare$
Thus, by Brokard's theorem it follows that triangle $ADB$ is self polar. Since $PP_1D \perp AB$ it must follow that $PP_1D$ is the diameter.
As such, $\measuredangle AQ_1B = \measuredangle APB = 90^\circ$ follows from Thale's theorem, and thus $AQ_1PB$ is cyclic.

Attachments:

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792 posts

#11 h Jan 9, 2024, 11:10 PM

Y by

Note that $PQQ_1P_1$ is cyclic, as
$\angle P_1QQ_1 = 90 - \angle BAQ = 90 - \angle CAP = \angle P_1PQ_1.$
Then Master Miquel tells us the Miquel point of $PQQ_1P_1$ is $P_1Q \cap AB = (P_1Q_1A) \cap (P_1PB)$ . Thus $\angle BPP_1 = \angle P_1Q_1A = 90$ , so $ABPQ_1$ is cyclic.

We finish by noting that $M$ is simply the midpoint of arc $PQ_1$ . $\blacksquare$

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bjump

#12 h Jul 23, 2024, 4:08 PM

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Claim: $PQQ_1 P_1$ is cyclic
Proof: $\angle P_1 P Q_1 = 90^{\circ} - \angle PAC = 90^{\circ} - \angle QAB = \angle P_1 Q Q_1$ which implies cyclicity. $\square$
Claim: $\angle BQ_1Q= \angle APQ = 90^\circ$
Proof: By Brokards the orthocenter of $A$ , $B$ , $Q_1P \cap P_1 Q$ is the center of $(PQQ_1 P_1)$ , recalling $P_1Q \perp AB$ means that the center lies on $P_1Q$ , which implies $P_1Q$ is a diameter which gives us the desired result. $\square$
Claim: $AQ_1 PB$ is cyclic
Proof: Note that $\angle AQ_1B= \angle BPA = 90^{\circ}$ . $\square$
Claim $M \in (AQ_1 PB)$
Proof: Let $X$ be the arc-midpoint of $PQ_1$ on $(AQ_1PB)$ not containing $A$ . Then $X\in MB, MA$ due to the fact $MA$ , and $MB$ bisect $\angle PAQ_1$ and $\angle PBQ_1$ thus $M=X \in (AQ_1 PB)$ and we are finished.

This post has been edited 1 time. Last edited by bjump, Jul 23, 2024, 4:08 PM

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bebebe

#13 h Aug 12, 2024, 11:51 PM

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Since $\angle PP_1Q=90-\angle BAP=90-\angle CAQ=\angle QQ_1P,$ we know $PP_1Q_1Q$ is cyclic. Lettin $X=P_1Q \cap PQ_1$ , by Brocard's Theorem, $polar(X)=AB.$ Since $X X' \perp AB$ and $P_1Q \perp AB,$ we know $X, X', P_1, Q$ are collinear. Since $O \in XX',$ we know $O \in P_1Q$ so $\angle P_1PQ=\angle P_1Q_1Q=90.$ Thus, $\angle P_1BP=180-\angle BPP_1 - \angle PP_1B=180-\angle AQ_1P_1 - \angle Q_1P_1A=\angle P_1AQ_1.$ Since $M$ bisects these angles, $AQ_1MPB$ is cyclic.

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872 posts

#14 h Aug 15, 2024, 3:03 AM

Y by

Notice that we don't need $B_1$ and $M$ , since $B_1$ does nothing, and if $AQ_1MPB$ is cyclic, $M$ is the midpoint of arc $PQ_1$ . Note how $\angle PQ_1Q=90-\angle CAQ$ and $\angle PP_1Q=90-\angle PAB$ . Since $\angle CAQ=\angle PAB$ , $PP_1Q_1Q$ is cyclic.

Now, if $X=PQ_1\cap P_1Q$ , then $AB$ is $X$ 's polar. Since $P_1Q\perp AB$ , this means that $P_1Q$ is a diameter of $(PP_1Q_1Q)$ . Therefore, $P_1$ is the orthocenter of $\triangle ABQ$ , and therefore $AQ_1PB$ is cyclic. $\square$

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#15 h Aug 21, 2024, 5:20 AM • 1 Y

Y by dolphinday

Note that the points $B_1$ and $M$ can be ignored, and so it suffices to prove that the points $A$ , $Q_1$ , $P$ , and $B$ are concyclic.

Claim: $PQQ_1P_1$ is a cyclic quadrilateral.

Proof: Notice that

$\angle P_1PQ_1 = 90^\circ - \angle PAC = 90^\circ - \angle BAQ = \angle P_1QQ_1. \ \square$

Suppose that $X = \overline{P_1Q} \cap \overline{PQ_1}$ . From Brokard, $\overline{AB}$ is the polar of $X$ . This means that $X$ lies on a diameter of $(PQQ_1P_1)$ , which implies $\overline{P_1Q}$ the desired diameter.

Now, we have

$\angle AQ_1B = \angle APB = 90^\circ$
due to Thales, which is what we want. $\square$

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#17 h Mar 27, 2025, 5:57 AM

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Let the feet from $P_1, Q_!$ to $AB, AC$ be $P_2, Q_2.$
Note that $P_1PQQ_1$ is cyclic because $\angle PP_1Q = \angle P_2P_1A = 90 - \angle P_1AB = 90 - \angle Q_1AC = \angle AQ_1Q_2 = \angle PQ_1Q.$
Let $X$ be the intersection $PQ_1, P_1Q.$
By Brokard’s, we have that the polar of $X$ with respect to $(P_1Q_1QP)$ is $AB.$ However, $P_1XQ$ is perpendicular to $AB.$ Thus, the center of $(P_1Q_1QP)$ lies on $P_1Q,$ impying that $\angle APB = 90.$
This finishes.

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