An isosceles triangle is given, prove a right angle

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geometry6

304 posts

geometry6

#1 h Jul 20, 2021, 9:20 PM • 9 Y

Y by centslordm, HWenslawski, mathematicsy, jhu08, megarnie, ImSh95, deplasmanyollari, Rounak_iitr, lian_the_noob12

Let $ABC$ be an isosceles triangle with $BC=CA$ , and let $D$ be a point inside side $AB$ such that $AD< DB$ . Let $P$ and $Q$ be two points inside sides $BC$ and $CA$ , respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$ . Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$ , and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$ , different from $C$ .
Suppose that $P$ , $E$ , $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$ .

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cadaeibf

700 posts

cadaeibf

#2 h Jul 20, 2021, 9:22 PM • 6 Y

Y by centslordm, jhu08, Pluto1708, megarnie, ImSh95, Jakjjdm

solution

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VulcanForge

626 posts

VulcanForge

#3 h Jul 20, 2021, 9:23 PM • 17 Y

Y by centslordm, Kagebaka, SatisfiedMagma, jhu08, Pluto1708, Steff9, CyclicISLscelesTrapezoid, rayfish, EpicBird08, ImSh95, axolotlx7, sabkx, shafikbara48593762, Ritwin, ihatemath123, Funcshun840, Jakjjdm

Let $M$ be the midpoint of $AB$ , so $CQPDM$ is cyclic with diameter $CD$ . Note $MP=MQ$ , since $CM$ bisects $\angle PCQ$ . Furthermore, the problem condition says $PQFC$ is an isosceles trapezoid. So $MC=MF$ , and thus since $M$ is the intersection of the perpendicular bisectors of $CF$ and $AB$ , it's the circumcenter.

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geometry6

304 posts

geometry6

#4 h Jul 20, 2021, 9:23 PM • 4 Y

Y by centslordm, jhu08, ImSh95, ItsBesi

Solution.Let $O'$ be the midpoint of $AB$ , we want to prove that $O'$ is the circumcenter of $(ABC)$ .

Since, $\angle{CPD}=\angle{CQD}=90^{\circ}\implies CQDP$ is cyclic. Also $\angle{CPD}=\angle{CO'D}=90^{\circ}\implies O'\in (CQDP)$ . Also, $CO'$ is the angle bisector of $\angle QCP\implies O'Q=O'P\implies O'$ lies on the perpendicular bisector of $QP$ .

$F$ lies on the perpendicular bisector of $QP\implies FQ=FP\implies \angle EPQ=\angle CQP\implies EQ=CP$ , also $CEQP$ is cyclic $\implies CEQP$ is isosceles trapezium $\implies$ the perpendicular bisectors of $QP$ and $CE$ are the same.

But now we have that $O'$ lies on the perpendicular bisector of $AB$ and $CE\implies O'$ is the circumcenter of $(ABC)\implies\angle{ACB}=90^{\circ}.\blacksquare$
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adj0109

17 posts

adj0109

#5 h Jul 20, 2021, 9:27 PM • 5 Y

Y by centslordm, jhu08, akasht, ImSh95, kinnikuma

Let $M$ be the midpoint of $AB$ and $l$ the perpendicular bisector of $PQ$

Claim 1. $M$ lies on $l$

proof. Since $CM \bot AB$ , $CQDMP$ is cyclic. Clearly $CM$ bisects $\angle QCP$ . Hence, $MQ = MP$ . $\blacksquare$

Claim 2. $l$ is also the perpendicular bisector of $CF$

proof. $\angle FCQ = \angle FPQ = \angle CQP \rightarrow CF || PQ$ Hence, PCFQ is an isosceles trapezium and proves the claim. $\blacksquare$

So, $M$ lies on the perpendicular bisector of both $AB$ and $CF$ which are both chords of $(ABC)$ , meaning $M$ is its center and $AB$ is a diameter. This proves $\angle ACB = 90^{\circ}$ .

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pad

1671 posts

pad

#6 h Jul 20, 2021, 11:28 PM • 3 Y

Y by centslordm, jhu08, ImSh95

Diagram
Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{PQ}$ respectively. Since $\angle CMD=90^\circ$ , and $\angle CPD=\angle CQD=90^\circ$ , we have $C,P,M,D,Q,F$ are concyclic, in particular all on $(CD)$ .

Lemma 1: $PM=QM$ .
Proof: We have
$\begin{align*} \angle PQM &= \angle PDM = \angle PDB = 90^\circ- \angle CBA \\ \angle QPM &= 180^\circ-\angle QDM = \angle QDA = 90^\circ - \angle CAB, \end{align*}$ which are equal since $\triangle CAB$ is isosceles. $\blacksquare$

Lemma 2: $CFQP$ is an isosceles trapezoid.
Proof: Since $P,E,F$ are collinear, we have
$\angle CQP=\angle EQP=\angle EPQ = \angle FPQ = \angle FCQ,$ so $\overline{CF}\parallel \overline{PQ}$ , proving the claim. $\blacksquare$

By Claim 2, the perpendicular bisector of $\overline{CF}$ and $\overline{PQ}$ are the same, so by Claim 1, $CM=FM$ . The circumcenter of $\triangle ABC$ is the intersection of the perpendicular bisectors of $\overline{CF}$ and $\overline{AB}$ , which is $M$ . Therefore, $\overline{AB}$ is a diameter of $(ABC)$ , so $\angle C=90^\circ$ .

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mathaddiction

308 posts

mathaddiction

#7 h Jul 20, 2021, 11:43 PM • 4 Y

Y by centslordm, TheMath_boy, jhu08, ImSh95

$[asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.92, xmax = 8.92, ymin = -7.52, ymax = 2.96; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-0.48,-2.03), 2.7600181158825747), linewidth(0.8) + ffvvqq); draw((-3.24,-2.02)--(2.28,-2.04), linewidth(0.8)); draw((2.28,-2.04)--(-0.47,0.73), linewidth(0.8)); draw((-3.24,-2.02)--(-0.47,0.73), linewidth(0.8)); draw(circle((-1.1650090578521075,-0.6474999671816959), 1.5429010824035494), linewidth(0.8) + ffvvqq); draw((-2.6820174749235983,-0.36602312512129576)--(0.21749097496619646,0.037509090670412214), linewidth(0.8) + linetype("4 4") + blue); /* dots and labels */ dot((-3.24,-2.02),dotstyle); label("$A$", (-3.8,-1.78), NE * labelscalefactor); dot((2.28,-2.04),dotstyle); label("$B$", (2.6,-1.86), NE * labelscalefactor); dot((-0.48,-2.03),linewidth(4pt) + dotstyle); label("$M$", (-0.4,-1.88), NE * labelscalefactor); dot((-0.47,0.73),linewidth(4pt) + dotstyle); label("$C$", (-0.74,1.1), NE * labelscalefactor); dot((-1.8600181157042153,-2.0249999343633904),dotstyle); label("$D$", (-1.78,-1.82), NE * labelscalefactor); dot((0.21749097496619646,0.037509090670412214),linewidth(4pt) + dotstyle); label("$P$", (0.3,0.2), NE * labelscalefactor); dot((-2.5475090906704123,-1.3325090250338023),linewidth(4pt) + dotstyle); label("$Q$", (-2.46,-1.18), NE * labelscalefactor); dot((-1.3933414138938967,-0.18667468888383276),linewidth(4pt) + dotstyle); label("$E$", (-1.32,-0.02), NE * labelscalefactor); dot((-2.6820174749235983,-0.36602312512129576),linewidth(4pt) + dotstyle); label("$F$", (-2.6,-0.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Let $M$ be the midpoint of $AB$ , obviously $D,M,P,C,F,Q$ are concyclic. Hence
$\angle FCA=\angle FCQ=\angle FPQ=\angle EPA=\angle EQP=90^{\circ}-\angle PQD=90^{\circ}-\angle DCB$ Therefore,
$\angle CFM=\angle CDM=90^{\circ}-\angle DCM=\angle ACM+\angle ACF=\angle FCM$ Hence $CM=FM$ , but $MA=MB$ , and $C,F,A,B$ are concyclic, hence $M$ is the center of the circle, which implies $\angle ACB=90^{\circ}$ .

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samrocksnature

8791 posts

samrocksnature

#8 h Jul 21, 2021, 12:00 AM • 2 Y

Y by jhu08, ImSh95

mathaddiction wrote:

$\angle FPQ=\angle EPA$

Why is this true?

Quote:

$\angle CDM=90^{\circ}-\angle DCM$

How do we know CMD=90?

Quote:

$90^{\circ}-\angle DCM=\angle ACM+\angle ACF$

pls explain

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IndoMathXdZ

691 posts

IndoMathXdZ

#9 h Jul 21, 2021, 12:29 AM • 2 Y

Y by jhu08, ImSh95

Let $\angle CBA = \theta$ .
Claim 01. $CP = FQ$ and $CQ = FP$ .
Proof. $CF \parallel PQ$ because $\measuredangle CQP = \measuredangle EQP = \measuredangle QPE = \measuredangle QPF = \measuredangle QCF$ . Therefore, $CPFQ$ is an isosceles trapezoid, forcing $CP = FQ$ and $CQ = FP$ .

Claim 02. $BP = CQ$ and $CP = QA$ .
Proof. Consider the spiral similarity $F : PQ \mapsto BA$ . Therefore,
$\frac{PB}{QA} = \frac{FP}{FQ} = \frac{CQ}{CP} \implies CP \cdot PB = CQ \cdot QA$ Since $BC = CA$ , then $BP + PC = QC + QA$ .
Consider the quadratic equation with $\{ CP, BP \}$ (this denotes length) as a root. Then,
$(x - BP)(x - CP) = x^2 - (BP + PC)x + (BP \cdot PC) = x^2 - (CQ + QA)x + (CQ \cdot QA) = (x - CQ)(x - QA)$ Thus, $\{ CP, PB \} = \{ QC, QA \}$ . However, $CP \not= CQ$ since this would force $D$ to be the midpoint of $AB$ as $DPCQ$ cyclic. Therefore, $CP = QA$ and $CQ = PB$ .

To finish this, let $BP = CQ = x$ and $CP = AQ = y$ . Then, $AB = (x + y) \cdot \cos \theta \cdot 2$ and $AB = BD + DA = \frac{y}{\cos \theta} + \frac{x}{\cos \theta}$ . Therefore, $\cos^2 \theta = \frac{1}{2}$ , which implies $\theta = 45^{\circ}$ as we know $\theta < 90^{\circ}$ .
This is enough to force $\angle ACB = 90^{\circ}$ .

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HamstPan38825

8857 posts

HamstPan38825

#10 h Jul 21, 2021, 1:19 AM • 2 Y

Y by jhu08, ImSh95

$[asy] size(250); pair A = origin, B = (8, 0), C = (4, 4); pair D = (5/2, 0); pair Q = foot(D, A, C), P = foot(D, B, C); pair EE = extension((A+B)/2, (P+Q)/2, A, C); pair X [] = intersectionpoints(circumcircle(A, B, C), circumcircle(C, P, Q)); pair F = X[1]; draw(A--B--C--cycle, red); draw(Q--D--P, magenta); draw(Q--F--C--P, green); draw(F--EE--P, lightcyan+dashed); draw(circumcircle(A, B, C), orange); draw(circumcircle(C, P, Q), gray); draw(C--D, brown+dashed); draw(P--Q, brown+dashed); pair X3 [] = intersectionpoints(C--D, P--Q); pair R = X3[0]; dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, S); dot("$Q$", Q, NW); dot("$P$", P, NE); dot("$F$", F, NW); dot("$E$", EE, SE); dot("$R$", R, SE); [/asy]$

First, observe that $P, E, F$ being collinear is equivalent to $CFQP$ being a trapezoid. This is just angle chasing: $\measuredangle FPQ = \measuredangle EPQ = \measuredangle PQE = \measuredangle PQC,$ from where $CFQP$ is an isosceles trapezoid because it is cyclic.

Next, we claim that this collinearity also implies that $\triangle FQA \sim \triangle FPB.$ This is again angle chasing: we have $\measuredangle AQF = \measuredangle CQF = \measuredangle CPF = \measuredangle BPF,$ and also $\measuredangle FAQ = \measuredangle FAC = \measuredangle FBC = \measuredangle FBP.$ In particular, this similarity implies that $\frac{FP}{PB} = \frac{FQ}{QA} \implies \frac{CQ}{PB} = \frac{CP}{QA} \implies \frac{CQ}{CP } =\frac{PB}{QA}$ From here, we present two finishes:

Finish 1 (Synthetic) Observe that $\frac{PB}{QA} = \frac{PD}{QD}$ because $\triangle AQD \sim \triangle BPD$ . We use quadrilateral $CQDP$ , which is trivially cyclic. Let the diagonals of the quadrilateral intersect at $R$ ; then the length condition implies $\frac{CQ}{PD} = \frac{QR}{PR} \text{ and } \frac{CP}{QD} = \frac{PR}{QR}.$ Combining these two conditions, we obtain $QR=PR$ ; similarly, $CR=RD$ so $\overline{PQ}$ is a diameter of $(APQ)$ . This implies the result. $\square$

Finish 2 (Trigonometric) Let $\angle CAB = \angle CBA = \theta$ , and moreover let $AC=BC=x$ . We have $\frac{PD}{QD} = \frac{BD}{AD}$ by similar triangles, while we can write $CQ=x-AQ = x-AD \cos \theta.$ Therefore, we have $\frac{BD}{AD} = \frac{x-AD \cos \theta}{x - BD \cos \theta}.$ This expands to $BDx - BD^2 \cos \theta = ADx - AD^2 \cos \theta \implies (BD-AD)x -\cos \theta(BD^2-AD^2) = 0.$ Factoring, we can write this as $(BD-AD)(x - \cos \theta (BD+AD)) = 0;$ but $BD+AD = \frac{x\sin(2\theta)}{\sin \theta}$ by the Law of Sines, so $x(BD-AD)\left(1 - \frac{\sin(2\theta)\cos \theta}{\sin \theta}\right) = 0,$ which implies that $\sin(2\theta) \cos(\theta) = \sin \theta \implies 2\sin \theta \cos^2 \theta = \sin \theta \implies \sin \theta(1-2\cos^2 \theta)=0.$ This tells us that $\cos^2 \theta = \frac 12$ , and since $\theta$ lies in the first quadrant, $\cos \theta = \frac{\sqrt 2}2$ , implying $\theta = 45^{\circ}$ . This finishes the problem. $\square$

Trigonometry is fun.

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rocketsri

976 posts

rocketsri

#11 h Jul 21, 2021, 1:45 AM • 2 Y

Y by jhu08, ImSh95

Let $M$ denote the midpoint of $AB$ , meaning $CM \perp AB$ . Since $PDQC$ is cyclic, we haver $\angle CPD = \angle CMD = 90$ , thus $M$ lies on the circumcircle of $PQDC$ , by inscribed angles. Note, this implies $\angle MCQ = \angle MCP$ as the angle bisector is the perpendicular bisector in an isosceles triangle, so the measure of arcs $MQ$ and $MP$ are equal, implying $MQ = MP$ . Then, note that $\angle FPQ = \angle CPQ$ so $FQPC$ is cyclic and for similar reasoning that $MQ = MP$ we have $FQ = PC$ , which means $FQPC$ is an isosceles trapezoid or $CF \parallel PQ$ . Thus, $M$ lies on the perpendicular bisectors of $CF$ and $AB$ and because they are both chords, the intersection of the perpendicular bisectors of chords is the cirumcenter by a famous lemma (Problem 1.40 of EGMO), and we are done. $\blacksquare$

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Pleaseletmewin

1574 posts

Pleaseletmewin

#12 h Jul 21, 2021, 2:34 AM • 2 Y

Y by jhu08, ImSh95

The collinearity implies that $PCFQ$ is an isosceles trapezoid.
Claim: $\triangle FQA$ and $\triangle FPB$ are isosceles.
Proof. Define $\angle CAB=\angle ABC=\theta$ and $\tfrac{1}{2}\widehat{FQ}=\tfrac{1}{2}\widehat{CP}=\alpha$ . Since $\angle QFP=\angle QFC=\angle ACB=180^\circ-2\theta$ and $\angle EAP=\angle EPQ=\alpha$ , we find $\angle FQE=2\theta-2\alpha$ . Now, observe that
$\begin{align*} \angle FAQ&=\angle FAC=\angle FAB-\angle CAB=180^\circ-\angle FCD-\angle CAB \\ &=180^\circ-(\alpha+(180^\circ-2\theta))-\theta=\theta-\alpha. \end{align*}$ Since this equals one-half of $\angle AQF$ 's supplement, the first claim is true. The second claim follows as the two triangles are similar.
Now, to finish, observe that
$\begin{align*} \frac{CD}{QC}=\frac{FQ}{FD}=\frac{QA}{DB}=\frac{QD}{DP}. \end{align*}$ This implies $QDPC$ is a rectangle, so we are done.
edit: dear god, how in the world did i miss perpendicular bisector...

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rocketsri

976 posts

rocketsri

#13 h Jul 21, 2021, 2:54 AM • 2 Y

Y by jhu08, ImSh95

samrocksnature wrote:

mathaddiction wrote:

$\angle FPQ=\angle EPA$

Why is this true?

Quote:

$\angle CDM=90^{\circ}-\angle DCM$

How do we know CMD=90?

Quote:

$90^{\circ}-\angle DCM=\angle ACM+\angle ACF$

pls explain

To the first one, he probably meant to write $\angle EPQ$ , to the second it's because the median is a perpendicular in an isosceles triangle.

To the third, it's a bit more complicated (there's probably an easier way but meh):

First, note that $\angle ACF + \angle ACM = \angle FCM$ . Also, $90 - \angle DCM = \angle CDM$ . Since $CM$ is an angle bisector, we have that $\angle QCM = \angle PCM$ , so we can subtract off those arcs. So, all that remains is to prove that $\angle FCQ = \angle CQP$ , which is true because $FCPQ$ is an isosceles trapezoid.

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DakuMangalSingh

72 posts

DakuMangalSingh

#14 h Jul 21, 2021, 3:31 AM • 1 Y

Y by jhu08

Let $M$ is the midpoint of $AB$ and $N$ is the midpoint of $PQ$ .

Since, $EQ=EP$ and $\angle PCQ=\angle QFP$ , so $FCQP$ is a cyclic trapezoid.

We have points $D,Q,C,P$ are concyclic.

Claim 1: Points $M,N,E$ are collinear.
Proof:Let the circumcircle of $\triangle CPQ$ intersects side $AB$ again at $M^\prime$ .Since $\angle CM^\prime D=90^\circ$ , so $M^\prime$ is the midpoint of $AB$ and also the midpoint of arc $PQ$ not containing $C$ , which implies $M^\prime$ lies on the perpendicular bisector of $PQ$ .
So, points $M,N,E$ are collinear.

Claim 2: $F$ is the center of spiral similarity taking $QA$ to $CB$ .
Proof: Trivial.

Claim 3: Points $E,Q,N,F$ are concyclic.
Proof: $\angle ENF=180^\circ-\angle FNM=180^\circ-\angle FQA=\angle EQF$
which implies points $E,Q,N,F$ are concyclic.

So, $\angle ACB=\angle QFP=180^\circ-\angle EFQ=180^\circ-\angle ENQ=90^\circ$

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MP8148

888 posts

MP8148

#15 h Jul 21, 2021, 4:23 AM • 1 Y

Y by jhu08

bruh

Let $M$ be the midpoint of $\overline{AB}$ , so that points $C$ , $D$ , $F$ , $P$ , $Q$ , $M$ lie on the circle with diameter $\overline{CD}$ .

Note that $MP = MQ$ . By symmetry, $CPQF$ is an isosceles trapezoid, so $MC = MF$ . Since $ABCF$ is cyclic, $MA=MB$ , and $MC=MF$ , we conclude the $M$ is the circumcenter.

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starchan

1601 posts

starchan

#68 h May 26, 2023, 8:10 AM • 1 Y

Y by ImSh95

ez complex
solution

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axolotlx7

133 posts

axolotlx7

#69 h Jun 17, 2023, 7:07 PM • 1 Y

Y by ImSh95

Let $M$ be the midpoint of $AB$ . Note that $\omega = (CPQDFM)$ with diameter $CD$ . As $CM$ bisects $\angle ACB$ , it bisects $\angle QCP$ and hence is the midpoint of arc $PQ$ of $\omega$ not containing $C$ . As $EP=EQ$ and $P,E,F$ collinear, we have that $E$ and $C$ are reflections across the perpendicular bisector of $PQ$ , which is $EM$ . Hence $FM=CM$ . Now $M$ is the intersection of perpendicular bisectors of $CF$ and $AB$ of $(ABCF)$ , so it must be the center of $(ABCF)$ . It follows that $\angle ACB = 90^{\circ}$ .

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eibc

597 posts

eibc

#71 h Jul 27, 2023, 2:54 AM

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Note that since
$\measuredangle FPQ = \measuredangle EPQ = \measuredangle PQE = \measuredangle PQC,$ quadrilateral $CPQF$ is an isosceles trapezoid.

Claim: $AQ = FQ$ .
Proof: Just angle chase:
$\begin{aligned} \measuredangle AFQ &= \measuredangle AQF + \measuredangle FAQ \\ &= \measuredangle CQF + \measuredangle FAC \\ &= \measuredangle CDF + \measuredangle FAC \\ &= (\measuredangle PDB + \measuredangle ADQ) + (\measuredangle CDP + \measuredangle QDF) + \measuredangle FAC \\ &= 2 \cdot \measuredangle CBA + 2 \cdot \measuredangle QCF + \measuredangle FAC \\ &= 2 \cdot (\measuredangle CBA + \measuredangle ABF) + \measuredangle FBC \\ &= \measuredangle CBF \\ &= \measuredangle QAF, \end{aligned}$ which gives what we wanted.

Now, note that we also have $FQ = CP$ from isosceles trapezoid $CPQF$ , so $AQ = CP$ . Thus, $QC = PB$ as well. To finish, denote $\alpha = \angle CAB = \angle ABC$ , and let $AD = x, BD = y$ . Straightforward computations gives $AQ = x \cos \alpha$ and $QC = PB = y \cos \alpha$ , as well as $AC = \tfrac{x + y}{2 \cos \alpha}$ . Then since $AQ + QC = AC$ we have $\tfrac{1}{2 \cos \alpha} = \cos \alpha$ or $\cos \alpha = \tfrac{\sqrt{2}}{2}$ , so $\alpha = 45^{\circ}$ and $\angle ACB = 90^{\circ}$ , as desired.

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cursed_tangent1434

557 posts

cursed_tangent1434

#74 h Sep 24, 2023, 1:49 PM • 1 Y

Y by Shreyasharma

OK, a slightly harder G1 ig.

We start off with the following observation.

Claim : Points $C,P,M,D,Q,F$ lie on the same circle with $CPQF$ infact being an isoceles trapezoid.

Proof : Note that $D$ clearly lies on $(CPQ)$ due to the right angles. Further, $CM \perp AB$ which gives $M$ also on $(CPQ$ . Further, we are given that $F$ also lies on this circle. Now, note that
$\measuredangle QCF = \measuredangle QPF = \measuredangle QPE = \measuredangle EQP = \measuredangle CQP$ Thus, $CF \parallel PQ$ which means $CPQF$ is indeed an isosceles trapezoid.

Let $O$ and $O_1$ be the centers of $(ABC)$ and $(CPQ)$ ). Let $M$ be the midpoint of $AB$ and $M_1$ the midpoint of $PQ$ . Now, we have the following key claim.

Claim : Points $E,O,O_1,M$ and $M_1$ lie on the same line.

Proof : Now, note that by the isosceles trapezoid, $CF\parallel PQ$ and we have $E-O_1-O$ . Clearly, the midpoint of $PQ$ also lies on its perpendicular bisector and thus $E-O_1-M_1-O$ . Now, also note that
$\measuredangle MPQ = \measuredangle MDQ = \measuredangle ADQ = \measuredangle PDB = \measuredangle PDM = \measuredangle PQM$ Thus, $MP=MQ$ . This means $M$ also lies on the perpendicular bisector of $PQ$ . Thus, $E-O_1-M_1-O-M$ . But, clearly $O, M$ also lie on the perpendicular bisector of $AB$ . This means both points $O$ and $M$ lie on the perpendicular bisector of $PQ$ and the perpendicular bisector of $AB$ . Since two distinct lines can intersect at most 1 point we have two possibilities. Either, $EO_1$ and $OM$ are the same line. But then, $E=C$ which is a degenerate case.
Or, we must have the two given intersection points being the same or $O=M$ . This happens if and only if $\angle ACB=90^\circ$ which was the desired conclusion.

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rafigamath

57 posts

rafigamath

#75 h Oct 31, 2023, 10:35 AM • 1 Y

Y by preparingfor_icho

By angle chasing,we get $PB=PF$ , and since $CFPQ$ is an isosceles trapezoid $PF=CQ$ ,so $PC=QA$ . $PC*PB=CQ*QA$ ,so $OP=OQ$ .This means $O$ and $O1$ are on the perpendicular bisector of $PQ$ ,and since $O1$ is also on the line $CD$ we get $PQ$ is diameter which gives us $\angle{PCQ}=90$ .

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Shreyasharma

666 posts

Shreyasharma

#76 h Feb 7, 2024, 12:47 AM

Y by

Note the spiral similarity from $F$ mapping $\overline{PQ} \mapsto \overline{BA}$ . Then from some simple angle chasing we may conclude,
$\begin{align*} \angle FCA &= \angle FBA\\ &= \angle FPQ\\ &= \angle EQP \end{align*}$ Then $\overline{PQ} \parallel \overline{CF}$ and $FCPQ$ is an isosceles trapezoid. Now let $M$ be the midpoint of $\overline{AB}$ which lies on $(FCPQD)$ as $ABC$ is isosceles. Now we claim $MP = MQ$ which is visible after some simple trig outlined below,
$\begin{align*} MQ &= \frac{AM \cdot \sin A }{\sin(\angle AQM)}\\ &= \frac{BC \cdot \sin B}{\sin(\angle AQD + \angle DQM)}\\ &= \frac{BC \cdot \sin B}{\sin(\angle BPD + \angle DPM)}\\ &= \frac{BC \cdot \sin B}{\sin(\angle MPB)}\\ &= MP \end{align*}$ Thus $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$ . However the $M$ is the intersection of two of the perpendicular bisectors of the cyclic quad $ABCF$ namely of $\overline{FC}$ and $\overline{AB}$ so $M$ is the center and thus $\angle ACB = 90$ . $\square$

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trk08

614 posts

trk08

#77 h Feb 10, 2024, 9:24 PM

Y by

Let $G$ be the midpoint of $AB$ . Note that $CPQD$ is cyclic.

Claim: $G$ lies on $(CPQ)$
Proof:
We angle chase:
$\angle CGD=90=\angle CQD$ $\square$

Claim: $GE$ is the perpendicular bisector of $PQ$
Proof:
Using LoS, we can say:
$GQ=\frac{\sin{\angle CAB}}{\sin{\angle AQG}}\cdot GA,$ $GP=\frac{\sin{\angle CBA}}{\sin{\angle GPC}}\cdot BG.$ By simple-angle chasing, the RHS of the above equations are the same, so $GP=GQ$ , as desired $\square$

Claim: $FCPQ$ is an isosceles trapezoid
Proof:
It is clear that:
$\angle EPQ=\angle EQP,$ so we are done by the fact that $FCPQ$ is cyclic $\square$

Therefore, $FC\parallel PQ$ , so $GE$ is the perpendicular bisector of $FC$ . Therefore, since $G$ is the midpoint of $AB$ , $G$ is the circumcenter of $(ABC)$ , so $\angle ACB=90$ $\blacksquare$ .

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AshAuktober

936 posts

AshAuktober

#78 h Jun 19, 2024, 3:10 AM • 1 Y

Y by GeoKing

Denote by $M$ the midpoint of $AB$ , and by $O$ the centre of the circumcircle of $\Delta ABC$ .
Claim 1: $MP = MQ$ .
Proof: Notice that $CQDMP$ is a cyclic polygon; as such, $\sin CPM = \sin BPM$ . But now the Law of Sines on triangles $AQM$ , $BPM$ finishes. $\square$

Claim 2: $CF \parallel PQ$ .
Proof: Since $P, E, F$ are collinear, $\angle CFP = \angle CQP = \angle FPQ$ (This last part is where we use $P, E, F$ collinear). $\square$

Note that due to Claim 2, for any point $X$ , $XC = XF \iff XP = XQ$ . But now $OC = OF \implies OP = OQ$ . Also $OA = OB$ .
But note that the conditions $XA = XB, XP = XQ$ are both satisfied by $O, M$ . So we break into two cases:

Case 1: $PQ \parallel AB$ (i.e. the perpendicular bisectors of $AB$ , $PQ$ coincide)
This isn't possible, as it would imply $AQ = BP$ , but $AQ/AD = BP/BD \implies AD = DB$ , contradiction.

Case 2: The aforementioned perpendicular bisectors do not coincide.
Then we get that they meet at either no point or at a unique point. But now $O, M$ lie on both, so we must have $O = M$ , i. e. The midpoint of $AB$ is the circumcentre of $\Delta ABC$ , giving us $\angle ACB = 90^\circ$ . $\square$

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giannis2006

45 posts

giannis2006

#79 h Jun 24, 2024, 2:15 PM

Y by

Let $M$ be the midpoint of $AB$ . $BC=CA=> \angle CMD = \angle CMA = 90 = \angle DPB = \angle DQA$ , hence $C,P,M,D,Q$ are concylic in a circle with diameter $CD$ . Hence, $F$ also lies on this circle. From the collinearity of $P,E,F$ we get (using $EQ=EP$ ) that: $\angle CFP = \angle CQP = \angle EQP = \angle EPQ = \angle FPQ => CF \parallel PQ$ and so $FCPQ$ is an isosceles trapezoid, as it is cyclic with $CF \parallel PQ$ , and hence $PQ,CF$ have the same perpendicular bisector. Now we have that: $BC=CA=> \angle ACM = \angle BCM => \angle QCM = \angle PCM => MQ=MP$ , as $QMPC$ is cyclic. Using that $PQ,CD$ have the same perpendicular bisector, we get that $MC=MF$ . Hence, $M$ is the point where the perpendicular bisectors of $AB,CF$ meet, so it is the center of the circle $(AFCB)$ , from where we conclude that $\angle ACB =90$ , as needed.

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MagicalToaster53

159 posts

MagicalToaster53

#80 h Sep 17, 2024, 12:16 AM • 1 Y

Y by MihaiT

After reading everyone else's solutions I can't help but think I made some error... Well it doesn't hurt to post...

We use directed angles $\measuredangle$ modulo $\pi$ . Let $M$ be the midpoint of $\overline{PQ}$ . Make the observation that $CPDQ$ is cyclic as $\measuredangle CQD = \measuredangle CPD = 90^{\circ}$ . Now observe $\triangle QEM \sim \triangle DCP.$ Indeed, we have that $\measuredangle MQE = \measuredangle PQC = \measuredangle PDC,$ and $\measuredangle EMQ = \measuredangle CPD = 90^{\circ}$ . We make the following claim:

Claim: $ECPM$ is cyclic.
Proof: We find that by angle chasing $\measuredangle QEM = \measuredangle DCP = \measuredangle MCP.$ However, $\measuredangle QEM = \measuredangle MEP,$ as $EM$ is the perpendicular bisector of $\overline{PQ}$ . $\square$

Hence we finish the problem by a final angle chase:
$\begin{align*} \measuredangle QDP &= \measuredangle QDC + \measuredangle CDP \\ &= \measuredangle QPC + \measuredangle CQP \\ &= \measuredangle QEM + \measuredangle EQM \\ &= 90^{\circ}. \end{align*}$ Therefore $\measuredangle QDP = \measuredangle QCP = \measuredangle ACB = 90^{\circ}$ , as desired. $\blacksquare$

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ezpotd

1251 posts

ezpotd

#81 h Sep 26, 2024, 5:47 AM

Y by

Since $C,E,Q$ and $P,E,F$ are collinear, we see that $CF$ is parallel to $PQ$ . Since $CF$ is the radical axis of $(ABC), (CPQ)$ , by LinPoP we know that the power difference is the same at $P,Q$ , and since $P,Q$ lie on $(CPQ)$ we just get that $AQ \cdot QC = CP \cdot PB$ . Let $M$ be the midpoint of $AB$ , then let $DM = x$ . Compute $AM = AC \cos BAC, AD = AC \cos BAC - x$ , symmetrically for $BM,BD$ . We then have that $(AC \cos BAC - x) (\cos BAC) (AC - (AC \cos BAC - x) \cos BAC) = (AC \cos BAC + x) (\cos BAC) (AC - (AC \cos BAC + x) \cos BAC)$ . This then forces $(AC \cos BAC - x)(AC \sin^2 BAC + x \cos BAC) = (AC \cos BAC + x)(AC \sin^2 BAC - x \cos BAC)$ , which resolves to $-\sin^2 BAC + \cos^2 BAC = \sin^2 BAC - \cos^2 BAC$ , which is just $\sin BAC = \cos BAC$ , forcing $\angle BAC = \frac{\pi}{4}$ as desired.

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Saucepan_man02

1299 posts

Saucepan_man02

#82 h Oct 31, 2024, 4:18 AM

Y by

Heres my solution:

Let $M$ denote the midpoint of $AB$ . Then, notice that: $M$ lies on $(CQDP)$ as $\angle CMB = \angle CQD = 90^\circ$ . Note that: $\angle BCM=\angle PCM = \angle ACM = \angle QCM$ which implies $MP=MQ$ . Thus, $MO \perp PQ$ . Also, note that: $\angle CQP=\angle FPQ$ which implies $CP=FQ$ or $CPQF$ is a cyclic trapezoid. Therefore, the perpendicular bisector of $CF$ passes through $M$ , which implies it's the circum-center of $\triangle ABC$ and we are done.

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ihatemath123

3440 posts

ihatemath123

#83 h Nov 26, 2024, 7:42 PM

Y by

Since $\angle FPQ = \angle CQP$ and $F$ lies on $(CPQ)$ , it follows that $\overline{CF} \parallel \overline{PQ}$ . Then, since $\angle CFD = 90^{\circ}$ , it follows that line $FD$ passes through the antipode of $C$ WRT $(ABC)$ – call this point $M$ . We also have from $\angle CFM = 90^{\circ}$ that line $DM$ is perpendicular to line $PQ$ . From this, we now have
$\begin{align*}\angle BDM &= \angle PDM - \angle PDB = \angle DPQ + \angle CBA - 180^{\circ} \\ &= \left( 180^{\circ} - 2 \angle CBA - \angle BCD \right) + \angle CBA - 180^{\circ} = \angle CDB, \end{align*}$ implying by symmetry (and the fact that $D$ is not the midpoint of $\overline{AB}$ ) that $\angle BCA = 90^{\circ}$ .

edit: wow #3 is beautiful. its a shame that you can solve the whole problem without constructing the midpoint of AB tho

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Ilikeminecraft

312 posts

Ilikeminecraft

#84 h Mar 13, 2025, 11:41 PM

Y by

Note that $\angle FPQ = \angle EPQ = \angle EQP = \angle CQP,$ so $CPQF$ is an isosceles trapezoid. Note that $MP= MQ$ since $CM$ bisects $\angle ACB.$ Thus, $CM = CF,$ but $MA = MB,$ so $M$ is circumcenter of $(ABC),$ finishing.

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bjump

994 posts

bjump

#85 h Mar 24, 2025, 10:06 PM

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By right angles we have $PQCDF$ cyclic. Now $EP=EQ$ because $E$ lies on the perpendicular bisector of $P$ , and $Q$ . So PoP gives $PE \cdot EF = CE \cdot EQ \implies CE=EF$ which means $CFQP$ is cyclic isosceles trapezoid. Note that the midpoint of $CD$ is the circumcenter of $(PQCDF)$ . And the midpoint of $AB$ lies on the circle by right angles, call it $H$ . $H$ lies on the perpendicular bisector of $CF$ , and $AB$ therefore it is the circumcenter of $(ABC)$ . Which implies that $\angle BCA = 90$ .

This post has been edited 1 time. Last edited by bjump, Mar 24, 2025, 10:06 PM

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