Segment has Length Equal to Circumradius

The post below has been deleted. Click to close.

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

djmathman

7934 posts

djmathman

#1 h Apr 30, 2014, 9:30 PM • 8 Y

Y by narutomath96, spacekid4, Davi-8191, centslordm, megarnie, Adventure10, Mango247, Rounak_iitr

Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$ . Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$ . Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$ .

This post has been edited 1 time. Last edited by djmathman, Apr 30, 2014, 9:56 PM

Z K Y

The post below has been deleted. Click to close.

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

BOGTRO

5818 posts

BOGTRO

#2 h Apr 30, 2014, 9:32 PM • 3 Y

Y by Condorcet, Adventure10, Mango247

I managed to reduce this to proving XY was perpendicular to BC, but i couldn't manage to prove that. Is this even true?

Z K Y

The post below has been deleted. Click to close.

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

AceOfDiamonds

1017 posts

AceOfDiamonds

#3 h Apr 30, 2014, 9:34 PM • 4 Y

Y by opptoinfinity, Imayormaynotknowcalculus, Adventure10, Mango247

Complex

Z K Y

The post below has been deleted. Click to close.

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

pi37

2079 posts

pi37

#4 h Apr 30, 2014, 9:37 PM • 9 Y

Y by mathandyou, narutomath96, r31415, ProbaBillity, yds, mathtiger6, Adventure10, Mango247, and 1 other user

Synthetic solution

Z K Y

The post below has been deleted. Click to close.

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

soy_un_chemisto

927 posts

soy_un_chemisto

#5 h Apr 30, 2014, 9:48 PM • 2 Y

Y by Adventure10, Mango247

djmathman wrote:

Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$ . Let $X$ be the circumcenter of triangle $AHC$ and $Y$ the orthocenter of triangle $APC$ . Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$ .

It should be:
let $P$ be the second intersection of the circumcircle of triangle $ABC$ with the internal bisector of the angle $\angle BAC$

EDIT: oops this is not the original problem but this version also has $XY$ equaling the circumradius.

This post has been edited 1 time. Last edited by soy_un_chemisto, Apr 30, 2014, 9:57 PM

Z K Y

The post below has been deleted. Click to close.

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

msinghal

725 posts

msinghal

#6 h Apr 30, 2014, 9:49 PM • 2 Y

Y by Adventure10, Mango247

Reflecting the whole thing over line AC gives a nice solution

Z K Y

The post below has been deleted. Click to close.

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

v_Enhance

6857 posts

v_Enhance

#7 h Apr 30, 2014, 9:59 PM • 12 Y

Y by dantx5, zschess, Ultroid999OCPN, Imayormaynotknowcalculus, HamstPan38825, megarnie, Kingsbane2139, nguyenducmanh2705, Adventure10, Mango247, Rounak_iitr, and 1 other user

msinghal wrote:

Reflecting the whole thing over line AC gives a nice solution

It also gives a nice complex bash

took me around 20 minutes, thankfully... I finished the test with five to spare so any more synthetic effort would have finished me.

Z K Y

The post below has been deleted. Click to close.

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

pi37

2079 posts

pi37

#8 h Apr 30, 2014, 10:25 PM • 3 Y

Y by megarnie, Adventure10, and 1 other user

Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)

Z K Y

The post below has been deleted. Click to close.

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

msinghal

725 posts

msinghal

#9 h Apr 30, 2014, 10:40 PM • 1 Y

Y by Adventure10

pi37 wrote:

Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)

I was thinking about that but I really didn't want to take care of that. I wasn't sure whether to add at the end that directed angles/lengths could resolve this, but decided against it.

Z K Y

The post below has been deleted. Click to close.

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

XmL

552 posts

XmL

#10 h Apr 30, 2014, 10:49 PM • 2 Y

Y by Adventure10, Mango247

Outline of mine

Z K Y

The post below has been deleted. Click to close.

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

JuanOrtiz

366 posts

JuanOrtiz

#11 h May 1, 2014, 12:42 AM • 2 Y

Y by Adventure10, Mango247

$Y$ is on circumcircle by angle chasing. Let $O_1$ be the reflection of $O$ through $AC$ . Then look at the rotohomothety with center $O$ that sends $AOY$ to $O_1OX$ (these triangles are similar). It turns out $AO_1=R$ and so $XY=R$ , so we're done.

hmm

This post has been edited 1 time. Last edited by JuanOrtiz, May 1, 2014, 1:17 AM

Z K Y

The post below has been deleted. Click to close.

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

henrypickle

238 posts

henrypickle

#12 h May 1, 2014, 12:59 AM • 1 Y

Y by Adventure10

Starting to worry that reducing to BC perpendicular to XY was not actually useful for finding a solution...

Z K Y

The post below has been deleted. Click to close.

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

JuanOrtiz

366 posts

JuanOrtiz

#13 h May 1, 2014, 1:16 AM • 2 Y

Y by Adventure10, Mango247

Well, it could be used to find the spiral similarity in my solution... I'm not sure.

Z K Y

The post below has been deleted. Click to close.

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

mathandyou

77 posts

mathandyou

#14 h May 1, 2014, 4:31 PM • 2 Y

Y by Adventure10, Mango247

pi37 wrote:

Synthetic solution

Why $(O_1)$ is the reflection of $(O)$ across $AC$ ?

Z K Y

The post below has been deleted. Click to close.

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

pohoatza

1145 posts

pohoatza

#15 h May 1, 2014, 6:15 PM • 3 Y

Y by HoRI_DA_GRe8, lrjr24, Adventure10

pi37 wrote:

Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)

Your solution was the same as the official solution, so it is safe to assume that you won't lose any points

.

This post has been edited 1 time. Last edited by pohoatza, May 1, 2014, 6:16 PM

Z K Y

The post below has been deleted. Click to close.

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

DrYouKnowWho

184 posts

DrYouKnowWho

#64 h Sep 24, 2021, 1:14 PM

Y by

Here's a walkthrough (outline) to my solution.

(a) Let $Y$ instead be a point on arc $ABC$ such that $YP\perp AC$
(b) Angle chase to confirm that $Y$ is actually the orthocenter of $\triangle APC$ , this also follows from the fact that $(AHC)$ is the reflection of $(ABC)$ over $AC$
(c) Angle chase to notice that $\triangle YSM_A$ is isoceles
(d) Let $X$ instead be the point such that $XOM_AY$ is a parallelogram, where $O$ is the center of $(ABC)$ , $M_A$ the midpoint of minor arc $BC$
(e) Angle chase to notice that $YM_A\perp AB$ , note that this also gives $X$ lies on the perpendicularbisector of $\overline{AB}$
(f) Drop perpendiculars from $X,Y$ to $AM_A$ , do some ratio chasing by introducing $L_A$ , the midpoint of arc $BAC$ .
(g) Conclude that $X$ also lies on the perpendicularbisector of $\overline{AP}$ , thus we have $X$ is the center of $(ABP)$ .

Z K Y

The post below has been deleted. Click to close.

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

Tafi_ak

309 posts

Tafi_ak

#66 h Dec 9, 2021, 12:32 AM

Y by

Something same to mathpirate
Let $AP$ intersect $(ABC)$ at $M$ . We claim that $Y$ lies on $(ABC)$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -4.5, xmax = 10., ymin = -10.5, ymax = 2.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw((0.8655617412366693,0.6216229568176963)--(-1.92,-4.78), linewidth(0.8)); draw((-1.92,-4.78)--(8.98,-4.76), linewidth(0.8)); draw((8.98,-4.76)--(0.8655617412366693,0.6216229568176963), linewidth(0.8)); draw(circle((3.5289007120376135,-4.17088806049879), 5.482840169518616), linewidth(0.8)); draw(circle((6.316661029199057,0.03251101731648583), 5.482840169518617), linewidth(0.8)); draw(circle((-0.9539487016641175,-1.8591284433689232), 3.076482693166482), linewidth(0.8)); draw((-0.9539487016641175,-1.8591284433689232)--(-0.9438884614073338,-7.341959383317089), linewidth(0.8) + ffxfqq); draw((0.8655617412366693,0.6216229568176963)--(4.710129680802359,1.1831977109320253), linewidth(0.8)); draw((4.710129680802359,1.1831977109320253)--(8.98,-4.76), linewidth(0.8)); draw((-0.9438884614073338,-7.341959383317089)--(8.98,-4.76), linewidth(0.8)); draw((-0.9438884614073338,-7.341959383317089)--(4.710129680802359,1.1831977109320253), linewidth(0.8)); draw((0.8655617412366693,0.6216229568176963)--(3.538960952294398,-9.653719000446957), linewidth(0.8)); draw(circle((-0.1282848169802912,-3.7524012016009203), 2.0654787792506077), linewidth(0.8)); draw(circle((2.3333146277043664,-6.489309204924442), 3.3863066036607337), linewidth(0.8)); draw((-0.9469953755773727,-5.648691160646276)--(1.8438718557541116,-3.1385603055018167), linewidth(0.8)); draw((-0.9539487016641175,-1.8591284433689232)--(-1.92,-4.78), linewidth(0.8)); draw((-1.92,-4.78)--(1.8438718557541116,-3.1385603055018167), linewidth(0.8)); draw((1.9702516157285461,1.0857410227694846)--(-1.92,-4.78), linewidth(0.8)); draw((-0.9539487016641175,-1.8591284433689232)--(3.5289007120376135,-4.17088806049879), linewidth(0.8) + ffxfqq); draw((3.5289007120376135,-4.17088806049879)--(3.538960952294398,-9.653719000446957), linewidth(0.8) + ffxfqq); draw((3.538960952294398,-9.653719000446957)--(-0.9438884614073338,-7.341959383317089), linewidth(0.8) + ffxfqq); /* dots and labels */ dot((0.8655617412366693,0.6216229568176963),dotstyle); label("$A$", (0.3729002823643257,1.0265932261440223), NE * labelscalefactor); dot((-1.92,-4.78),dotstyle); label("$B$", (-2.4152150566409243,-5.7012503092816855), NE * labelscalefactor); dot((8.98,-4.76),dotstyle); label("$C$", (9.383765798279844,-5.054730810381917), NE * labelscalefactor); dot((3.5289007120376135,-4.17088806049879),dotstyle); label("$O$", (3.6054977768631664,-3.9637291559885597), NE * labelscalefactor); dot((0.8677603171614444,-0.576600922184725),dotstyle); label("$H$", (0.5143264227486499,-1.3776511603894888), NE * labelscalefactor); dot((1.8438718557541116,-3.1385603055018167),dotstyle); label("$P$", (3.1004044183477224,-2.670690158189024), NE * labelscalefactor); dot((-0.9438884614073338,-7.341959383317089),dotstyle); label("$Y$", (-2.3141963849378357,-7.2569378535092515), NE * labelscalefactor); dot((-0.9539487016641175,-1.8591284433689232),dotstyle); label("$X$", (-1.122176058841388,-1.5796885037956663), NE * labelscalefactor); dot((3.538960952294398,-9.653719000446957),dotstyle); label("$M$", (3.5852940425225484,-10.186479332898823), NE * labelscalefactor); dot((4.710129680802359,1.1831977109320253),dotstyle); label("$P'$", (4.83792557164085,1.5518903190000835), NE * labelscalefactor); dot((-0.9469953755773727,-5.648691160646276),dotstyle); label("$R$", (-0.718101372029033,-6.246751136478364), NE * labelscalefactor); dot((-0.9485925747692824,-4.778217601054623),dotstyle); label("$S$", (-0.8595275124133572,-4.569841186207092), NE * labelscalefactor); dot((1.9702516157285461,1.0857410227694846),dotstyle); label("$H'$", (2.049810232635599,1.2892417725720529), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Note that
$\begin{eqnarray*} \measuredangle PAC+\measuredangle PCA=\measuredangle P'AB+\measuredangle CBP'=\measuredangle CBA \end{eqnarray*}$ Which means $Y$ lies on $(ABC)$ .

Clearly $OM\perp BC$ and $OX\perp AB$ . Now $\measuredangle MYC=\measuredangle MAC=\measuredangle PHA+90^\circ -\measuredangle CBA=\measuredangle HCP+\measuredangle PCY=\measuredangle ACY$ Which means $MY\perp AB$ . Now we if we can prove that $XY\perp BC$ then we are basically done. Because then $MYXO$ will be a parallelogram and we get our desired result.

Now suppose that $R=(OBX)\cap MYP$ . Then an simple angle chase gives that $\measuredangle PRY=\measuredangle PMY=\measuredangle AP'P=\measuredangle PBX=\measuredangle PRX$ Therefore $X,R,X$ collinear and similarly $B,R,M$ collinear. And another angle chasing gives that $\measuredangle SRB+\measuredangle SBR=\measuredangle XPB+\measuredangle MBC=90^\circ$ And we are done. $\square$

Z K Y

The post below has been deleted. Click to close.

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

Mogmog8

1080 posts

Mogmog8

#68 h Apr 22, 2022, 5:08 AM • 1 Y

Y by centslordm

Let $O_1$ be the center of $(AHC).$ Notice $(ABC)$ and $(AHC)$ are reflections over $\overline{AC},$ and since the reflection of $Y$ over $\overline{AC}$ lies on $(AHC),$ we see $Y$ lies on $(ABC).$ Notice $\overline{AB}\perp\overline{OX},\overline{AP}\perp\overline{O_1X},$ and $\overline{AB}\perp\overline{OO_1}.$ Hence, $\measuredangle XO_1O=90-\measuredangle(\overline{XO_1},\overline{AC})=\measuredangle PAC=\measuredangle BAP=90-\measuredangle(\overline{AP},\overline{XO})=\measuredangle OXO_1.$ But $\angle AOY=2\angle ADY=180-2\angle DYC=180-\angle A=\angle O_1OX$ so $\triangle OAY\sim\triangle OO_1X$ and $O$ is the center of the spiral similarity $\overline{XY}\mapsto\overline{O_1A}.$ Therefore, $\triangle AOO_1\cong\triangle YOX.$ $\square$

Z K Y

The post below has been deleted. Click to close.

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

Number1048576

91 posts

Number1048576

#69 h Jul 7, 2022, 8:15 PM • 1 Y

Y by Mango247

hint 1
hint 2
hint 3
hint 4
hint 5
solution

Z K Y

The post below has been deleted. Click to close.

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

ike.chen

1162 posts

ike.chen

#70 h Aug 23, 2022, 12:04 PM

Y by

Let the midpoints of $AB$ and $AC$ be $M$ and $N$ respectively, the circumcenter of $ABC$ be $O$ , the reflections of $O$ and $Y$ over $AC$ be $O_B$ and $Y_1$ respectively, $K = AP \cap OM$ , and $T = AP \cap O_BX$ .

The Orthocenter Reflection Lemma implies $(ABC)$ and $(AHC)$ are symmetric about $AC$ , so $O_B$ is the center of $(AHC)$ . Now, because the aforementioned lemma yields $Y_1 \in (AHPC)$ , we have $Y \in (ABC)$ .

Since $AP$ is the common chord of $(APB)$ and $(AHPC)$ , we know $AP \perp O_BX$ . Thus, because $XA = XB$ gives $X \in OM$ , $\angle XOO_B = \angle MON = 180^{\circ} - \angle A$ and $\angle OXO_B = \angle KXT = 90^{\circ} - \angle XKT = \angle MAK = \frac{\angle A}{2}$ which means $OO_B = OX$ .

Let $l$ be the perpendicular bisector of $O_BX$ . Now, since $OC = OY$ and $CY \perp AP \perp O_BX$ follows from the orthocenter condition, we know $C$ and $Y$ are symmetric about $l$ . Hence, reflection properties imply $XY = O_BC = OC$ as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

AwesomeYRY

579 posts

AwesomeYRY

#71 h Mar 21, 2023, 2:23 PM

Y by

Claim 1: $Y\in (ABC)$
Proof: Note that the circle $(AHC)$ is the reflection of $(ABC)$ over line $AC$ . Thus, $P'$ , the reflection of $P$ over $AC$ , lies on $(ABC)$ . Since $P$ is the orthocenter of $AYC$ , we also have that $Y,A,C,P'$ is cycle, which finally shows that $Y\in (ACP') =(ABC)$ $\square$

Claim 2: $\triangle OXO_1$ is isosceles
Proof: Note that
$\angle (OX,XO_1) = \angle (AB,AP) = \angle (AP,AC) = \angle (O_1X, OO_1)$ due to perpendicularity conditions $\square$

Claim 3: $YXO_1C$ is an isosceles trapezoid.
Proof: Clearly, $XO_1\perp AP \perp YC\Longrightarrow XO_1\parallel YC$ . Additionally, the perpendicular bisectors both pass through $O$ and are parallel, so they must be the same line, and therefore it is an isosceles trapezoid. $\square$

Thus, $XY = O_1C = R$ and we're done. $\blacksquare$ .

Z K Y

The post below has been deleted. Click to close.

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

pikapika007

294 posts

pikapika007

#72 h Jul 1, 2023, 4:41 PM

Y by

this problem is so boring

Let $M$ be the midpoint of arc $BC$ in $(ABC)$ , and let $O$ be the circumcenter of $ABC$ . We prove that $XYMO$ is a parallelogram, which finishes.
Note that $ABYC$ is cyclic since $\angle ABC = 180 - \angle APC = \angle AYC$ .
Claim: $\overline{XO} \parallel \overline{MY}$ .

Proof. Note that $\overline{XO} \perp \overline{BC}$ , so it is enough to show that $\overline{YM} \perp \overline{BC}$ . Indeed, $\angle BAD = \frac{\angle A}{2}$ , while $\angle YDA = \angle YCA = 90 - \frac{\angle A}{2}$ since $P$ is the orthocenter in $AYC$ . Since $\angle BAD + \angle YDA = 90$ , we are done. $\square$

Claim: $\overline{XY} \perp \overline{BC}$ .

Proof. Note that $\angle PYM = \angle PYC + \angle CYD = \angle A$ , and $\angle PXB=2\angle PAB=\angle A$ . Also note that $M$ is the reflection of $P$ over $YC$ , hence $PYM$ is isosceles. Since $PXB$ is also isosceles, in fact we have $PXB \sim PYM$ . Now by spiral similarity, $\overline{BM} \cap \overline{XY}$ is on $(PXB)$ , $(PYC)$ and hence their angle between them is equal to $\angle XPB = 90 - \frac{\angle A}{2}$ . Also, $\angle MBC = \frac{A}{2}$ and hence $\overline{XY} \perp \overline{BC}$ . $\square$

To finish, note that $\overline{OM} \perp \overline{BC}$ and we are done since $XY$ is parallel to $OM$ . $\blacksquare$

This post has been edited 3 times. Last edited by pikapika007, Jul 2, 2023, 1:08 AM

Z K Y

The post below has been deleted. Click to close.

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

ihatemath123

3427 posts

ihatemath123

#73 h Jul 2, 2023, 1:03 AM

Y by

Due to orthocenter reflecting stuff, the circumcircle of $APHC$ is the reflection of the circumcircle of $\triangle ABC$ about $\overline{AC}$ . Let $O_1$ and $O_2$ be the centers of $(ABC)$ and $(APC)$ , respectively.

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AP}$ , respectively. Then, clearly, $X$ is the intersection of rays $O_1 M$ and $O_2 N$ , since those are the perpendicular bisectors of $\overline{AB}$ and $\overline{AP}$ .

Since $\angle XMA = \angle XNA = 90^{\circ}$ , it follows that $XMNA$ is cyclic. Letting $T$ be the intersection of $\overline{AC}$ and $\overline{O_1 O_2}$ , we have $\angle ATO_2 = 90^{\circ}$ due to symmetry, so $\angle ANO_2 = \angle ATO_2 = 90^{\circ}$ , so $ANTO_2$ is cyclic. Thus,
$\angle O_1X O_2 = \angle MXN = \angle MAN = \angle CAN = \angle TAN = \angle TO_2 N = \angle O_1 O_2 X$ so $\triangle O_1 X O_2$ is isosceles with $O_1 X = O_1 O_2$ and $\angle O_1XO_2 = \angle O_1 O_2 X = \frac{\angle A}{2}$ .

Now, we claim that $\triangle O_1 YX \cong \triangle O_2 A O_1$ .

A tiny bit of angle chasing gives us that $\angle YO_1 X = \angle AO_1 O_2$ . Since $YO_1 = AO_2$ , $O_1 X = O_2 O_1$ and $\angle YO_1 X = \angle AO_1 O_2$ , we have $\triangle YO_1 X \cong \triangle AO_1 O_2$ by SAS congruence. Thus, $XY = AO_2 = AO_1 = R$ .

This post has been edited 1 time. Last edited by ihatemath123, Jul 2, 2023, 1:31 AM

Z K Y

The post below has been deleted. Click to close.

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

joshualiu315

2513 posts

joshualiu315

#74 h Feb 27, 2024, 6:12 AM

Y by

this problem is middle of the line imo

not too easy, not too hard, not too bashy, not too clean

Let the center of $(ABC)$ be $O$ and the center of $(AHC)$ be $O'$ . It is well known that $(ABC)$ and $(AHC)$ are reflections of each other, so $AO=AO'$ .

Note that

$\angle AYC = 180^\circ - \angle APC = 180^\circ - \angle AHC = \angle ABC,$
hence $Y$ lies on $(ABC)$ .

Also, $\overline{OX}$ and $\overline{OO'}$ are perpendicular to $\overline{AB}$ and $\overline{AC}$ , respectively. Thus letting $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ give that $AMON$ is cyclic, which means $\angle MON = \angle XOO' = 180^\circ - \angle BAC$ .

Then, notice

$\angle AOY = 2 \angle ACY = 2 (90^\circ-\angle PAC) = 180^\circ - 2 \angle PAC = 180^\circ - \angle XOO'.$
This implies that rays $OO'$ and $OX$ can be rotated around $O$ to coincide with rays $OA$ and $OY$ , respectively. Hence, we can rotate rays $OO'$ and $OA$ around $O$ to coincide with rays $OX$ and $OY$ , giving $\angle AOO' = \angle XOY$ .

Finally, realize that

$\angle OXO' = \angle BAP = \angle CAP = \angle O'XO,$
so $\triangle OXO'$ is isosceles with $OX=OX'$ . Combine this with the aforementioned angle condition and the fact that $OA=OY$ , we get that $\triangle OXY \cong \triangle OO'A$ ; therefore, $XY=OA'=OA$ . $\square$

Z K Y

The post below has been deleted. Click to close.

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

Inconsistent

1455 posts

Inconsistent

#75 h Mar 5, 2024, 7:32 PM

Y by

Quite a fun problem! Trig bash ftw.

First, notice that $\angle B = 180^{\circ} - \angle AHC = 180^{\circ}-\angle APC = \angle AYC$ so $(ABYC)$ cyclic. Now note that $X$ is simply the intersection of the perpendicular bisector of $AB$ with the perpendicular bisector of $AP$ . In particular, $AP \perp YC$ . Let $O'$ be the center of $(AHC)$ and the reflection of $O$ over $AC$ , and let $M$ be the midpoint of $AC$ . It follows that the projection in the direction of $CY$ from the three points $O, M, O'$ map to three points $O, X', X$ along the perpendicular bisector of $AB$ , such that $X'$ is the midpoint of $OX$ . It suffices to show that $XY = YO = R$ , so we simply need to show that $YX' \perp OX \perp AB$ , which is equivalent to $d(X, AB) = d(Y, AB)$ .

Now, we begin the trig bash. Let $M'$ be the midpoint of $AB$ . Then by LoS in $\triangle MM'X$ , we have $d(X, AB) = \sin \angle M'MX \cdot \frac{M'M}{\sin \angle M'XM} = \sin \left( \frac{B - C}{2} \right) \cdot R \cdot \frac{\sin A}{\sin \frac{A}{2}}$

Now by generalized LoS we have $d(Y, AB) = BY \cdot \sin \angle ABY = 2R \cdot \sin \left( \frac{B - C}{2} \right) \cdot \cos \frac{A}{2}$ .

Thus equating the two, it is equivalent to $\sin A = 2\sin \frac{A}{2} \cos \frac{A}{2}$ , which follows from double angle.

Z K Y

The post below has been deleted. Click to close.

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

HamstPan38825

8847 posts

HamstPan38825

#77 h Mar 13, 2024, 3:26 AM

Y by

Here's a complex solution very similar to Evan's (even down to the point names!)

We will instead introduce the point $Q$ , the reflection of $P$ over $\overline{AC}$ , which lies on $(ABC)$ . WLOG assume $(ABC)$ has unit radius. Observe that the bisector condition implies that $q^2 = \frac{c^3}b$ (say by considering the arc midpoint). Now, let $x$ and $y$ represent the complex numbers associated with the reflections of $X$ and $Y$ over $\overline{AC}$ . We have $y = a+q+c$ , and $x = a+\frac{-(q-a)(b'-a)(\overline{q-a} - \overline{b'-a})}{(q-a)(\overline{b'-a}) - (\overline{q-a})(b'-a)}$ by the circumcenter formula, where $b'$ is the reflection of $B$ over $\overline{AC}$ . It follows that
$\begin{align*} x-y &= q+c+\frac{-(q-a)\left(\frac 1c-\frac b{ac}\right)-\left(\frac 1q - \frac 1a\right)\left(c-\frac{ac}b\right)}{(q-a)\left(\frac 1c-\frac b{ac}\right) - \left(\frac 1q-\frac 1a\right)\left(c-\frac{ac}b\right)} \\ &= q+c+\frac{c(q-a)\left(\frac{ac}q-c-a+b\right)}{bq-ab+\frac{ac^2}q-c^2} \\ &= q+c + \frac{c(q-a)\left(\frac{ac}q-c-a+\frac{c^3}{q^2}\right)}{\frac{c^3}q-\frac{ac^3}{q^2}+\frac{ac^2}q - c^2} \\ &= q+c+\frac{c(q-a)(acq-cq^2-aq^2+c^3)}{c^2(c-q)(q-c)} \\ &= q+c-\frac{aq+c^2+cq}c \\ &= -\frac{aq}c. \end{align*}$ This clearly has magnitude $1$ , as needed.

This post has been edited 2 times. Last edited by HamstPan38825, Mar 13, 2024, 3:27 AM

Z K Y

The post below has been deleted. Click to close.

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

OronSH

1718 posts

OronSH

#78 h Dec 25, 2024, 6:02 PM • 1 Y

Y by Zhaom

Let $N$ be the minor arc midpoint. First notice that $Y\in(ABC)$ since $\angle ABC=180-\angle AHC=180-\angle APC=\angle AYC$ .

Next $\angle CNP=\angle CBA=180-\angle CHA=180-\angle CPA=\angle CPN$ , and $CY\perp AP$ , so $CY$ perpendicularly bisects $PN$ .

Next if $Z=CY\cap AN$ then $\measuredangle ZYN=\measuredangle ZAB$ implies $(AZY)$ passes through $AB\cap YN$ , so $AB\perp YN$ and thus $XO\parallel YN$ , where $O$ is the circumcenter.

Finally notice that the projections from $X,O$ to $AN$ have distance $\tfrac12PN=ZN$ , and thus $XO\parallel YN$ implies $XO=YN$ . Thus $XONY$ is a parallelogram and $XY=ON$ as desired.

Z K Y

The post below has been deleted. Click to close.

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

Mathandski

706 posts

Mathandski

#79 h Mar 7, 2025, 10:07 PM • 2 Y

Y by OronSH, Zhaom

Angles directed by default.

Note that $(AHPC)$ is the reflection of $(ABC)$ over $AC$ and $(AYC)$ is the reflection of $(APC)$ over $AC$ so $(ABYC)$ is cyclic. We instead define $Y$ as the second intersection of the line through $C$ perpendicular to $AM$ with $(ABC)$ .

Claim 1: $OX \parallel MY$ .

We check that $MY \perp AB$ to suffice. We angle chase to get $\angle BMY_1 = \angle BCY_1 = C + \frac A2 - 90$ and $\angle MBA = B + \frac A2$ as desired. $\Box$

Claim 2: $XY \parallel OM$ .

Note that $\angle PXB = 2\angle PAB = \angle A$ and $\angle PYM = \angle (PY, YM) = \angle (CA, AB) = \angle A$ . Furthermore, $\angle PBX = 90 - \angle A2$ from $PX = XB$ and $\angle PMY = \angle ACY = 90 - \angle A2$ as well meaning there is a spiral similarity at $P$ sending $MY$ to $BX$ . Analogously, there is a spiral similarity from $BM$ to $YX$ .
$\angle (BC, XY) = \angle BCM + \angle (BM, XY)$ $\angle \frac A2 + \angle (PM, PY) = \angle \frac A2 + 90 - \angle (PM, AC) = 90$
Therefore, $OM \perp BC \perp XY$ as desired $\Box$ .

With both claims proven, we have $OXYM$ is a parallelogram meaning $XY = OM = R$ as desired.

Z K Y

The post below has been deleted. Click to close.

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

ihatemath123

3427 posts

ihatemath123

#80 h Mar 12, 2025, 2:34 AM • 2 Y

Y by OronSH, Zhaom

Let $O$ be the circumcenter. Let $X'$ be the reflection of $X$ across the perpendicular bisector of line $CY$ . (Note that line $CY$ is perpendicular to bisector of $\angle BAC$ .) Since $\overline{OX} \perp \overline{AB}$ , we have $\overline{OX'} \perp \overline{AC}$ . So, $X'$ lies on the perpendicular bisector of $\overline{AC}$ . Also, since $X$ lies on the perpendicular bisector of $\overline{AP}$ , so does $X'$ . Therefore, $X'$ is the circumcenter of $\triangle APC$ . We have $XY = X'C$ , and $X'C$ is obviously equal to the circumradius.

Z K Y

The post below has been deleted. Click to close.

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

Zhaom

5123 posts

Zhaom

#81 h Yesterday at 9:28 PM

Y by

Note that $Y$ lies on $(ABC)$ . Let $K$ be the midpoint of arc $\overarc{CAB}$ . It suffices that $XKOY$ is a rhombus. Let $X'$ be the point such that $X'KOY$ is a rhombus. First, note that $\angle{}AKY=\angle{}ACY=90^\circ-\tfrac{\angle{}CAB}{2}=180^\circ-\angle{}BAK$ , so $\overline{KY}\parallel\overline{AB}$ , so $X'$ lies on the perpendicular bisector of both $\overline{KY}$ and $\overline{AB}$ . Now, we see by linearity of power of a point that since if $O'$ is the point such that $OCO'A$ is a parallelogram then $O'$ lies on the perpendicular bisector of $\overline{AP}$ , as $\overline{CY}\perp\overline{AP}$ , we have that
$\begin{align*} X'A^2-X'P^2&=\left(KA^2-KP^2\right)+\left(YA^2-YP^2\right)-\left(OA^2-OP^2\right)\\ &=\left(KA^2-KP^2\right)+\left(CA^2-CP^2\right)-\left(\left(AA^2-AP^2\right)+\left(CA^2-CP^2\right)-\left(O'A^2-O'P^2\right)\right)\\ &=KA^2-KP^2+AP^2\\ &=0 \end{align*}$ by the Pythagorean theorem, so $X'$ lies on the perpendicular bisector of $\overline{AP}$ and we are done.

This post has been edited 1 time. Last edited by Zhaom, Yesterday at 9:28 PM

Z K Y