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most bruh geo you can imagineeeeeeeeeeeeee
ItzsleepyXD   1
N 11 minutes ago by Funcshun840
Source: bruhhhhhh
Let $ABC$ be triangle with $AC>AB$ and $B'$ on the segment $AC$ such that $AB=AB'$ . Consider point $E,F$ on $AB,AC$ such that $EF \parallel BB'$ . Point $T$ is the intersection of tangent line through point $E,F$ to circle $(EBC),(FBC)$ respectively . If the tangent through point $B'$ to $(BB'C)$ intersect $AB$ at $K$ . Line $KT$ intersect $BC$ at $D$ . Prove that $AD$ bisect $\angle BAC$ .
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ItzsleepyXD
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Funcshun840
11 minutes ago
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Collinearity with orthocenter
liberator   182
N May 19, 2025 by CrazyInMath
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
182 replies
liberator
Jan 4, 2016
CrazyInMath
May 19, 2025
Collinearity with orthocenter
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G H BBookmark kLocked kLocked NReply
Source: IMO 2013 Problem 4
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fasttrust_12-mn
118 posts
#186
Y by
let $\omega_3$ be the circumcircle of $\triangle ANM$ donate $E$ be the miquel point now we have that $A-E-W$ as a fast result we get $X-E-H-Y$



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[/asy]
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MagicalToaster53
159 posts
#187
Y by
We use directed angles $\measuredangle$ modulo $\pi$. Let $Z$ be the second point of intersection between $\omega_1$ and $\omega_2$. Then observe that $X, Z, Y$ are collinear as $\measuredangle XZW = \measuredangle YZW = 90^{\circ}$. Let $D$ be the point where the altitude from $A$ meets $\overline{BC}$. Then make the observation that $BX \parallel HD$, as $\measuredangle XBW = \measuredangle 90^{\circ} = \measuredangle HDW$. We now make a claim:

Claim: $X, H, Z$ are collinear.
Proof: We have the following angle chase: \[\measuredangle BXH = \measuredangle DHZ = \measuredangle DWZ = \measuredangle BWZ = \measuredangle BXZ. \square\]
Hence as $XZ$ is the same line passing simultaneously through both $H$ and $Y$, we find that $X, H, Z, Y$ are collinear, as desired. $\blacksquare$
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ezpotd
1287 posts
#188
Y by
Draw in $(BWN) \cap (CWM) = K$, then Miquel tells us $(AMNHK)$ is cyclic. Direct angles. We then see $\angle NKX = \angle NBX = 90 - \angle ABC = \angle NAH = \angle NKH$, so $K,H,X$ are collinear, and so are $K,H,Y$, finishing.
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legogubbe
19 posts
#189
Y by
Let $Z$ be the second intersection point of $(BWN)$ and $(CWN)$.

Claim 1. Points $X$, $Y$ and $Z$ are collinear.
Proof. We see by constructions that $BXZW$ and $CYZW$ are cyclic. This implies
\[ \measuredangle WZX = \measuredangle WBX = 90^{\circ} = \measuredangle WCY = \measuredangle WZY, \]which means $X$, $Y$ and $Z$ are indeed collinear.

Claim 2. Quadrilateral $AHZM$ is cyclic.
Proof. By Miquel's theorem, $ANZM$ is cyclic. However, by the properties of the orthocenter, $ANHM$ is cyclic. From this we deduce that in fact, $ANHZM$ is cyclic, thereby $AHZM$ is cyclic.

Let $K$ be the intersection point of lines $AH$ and $BC$.

Claim 3. Quadrilateral $KHZW$ is cyclic.
Proof. From cyclic quadrilaterals,
\[ \measuredangle KHZ = \measuredangle AHZ = \measuredangle AMZ = \measuredangle CMZ = \measuredangle CWZ = \measuredangle KWZ, \]proving our claim.

Finally, $\measuredangle WZH = \measuredangle WKH = \measuredangle CKA = 90^{\circ}$. Hence $X$, $Y$, $Z$ and $H$ are collinear.
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cursed_tangent1434
640 posts
#190
Y by
We let $T$ denote the second intersection of circles $(BNW)$ and $(CMW)$. We start ff by noting that since,
\[\measuredangle WTX = \measuredangle WBX = \frac{\pi}{2}\]and
\[\measuredangle WTY = \measuredangle WCY = \frac{\pi}{2}\]which implies that points $X$ , $T$ and $Y$ are collinear. Next, by Radical Center Theorem on $(BNTW)$ , $(CMTW)$ and $(BNMC)$, lines $\overline{BN}$ , $\overline{TW}$ and $\overline{MC}$ concur. It is clear that $A= BN \cap CM$, so
\[AT \cdot AW = AN \cdot AB = AH \cdot AD\]which implies that $DHTW$ is also cyclic. Now this finishes since,
\[\measuredangle WTH = \measuredangle WDH = \frac{\pi}{2} = \measuredangle WTX\]which implies that $X$ , $H$ and $T$ are collinear. Combining this with our previous observation concludes that points $X$ , $Y$ and $H$ are collinear, as desired.
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Saucepan_man02
1358 posts
#191
Y by
Storage
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cj13609517288
1922 posts
#192 • 1 Y
Y by peace09
Let $Z$ be the other intersection of $\omega_1$ and $\omega_2$. By radax on $\omega_1,\omega_2,(BNMC)$ we get that $A,Z,W$ are collinear. Since $\angle XZW=\angle YZW=90^{\circ}$, we get that $X,Y,Z$ are collinear. Thus it suffices to show that $\angle HZW=90^{\circ}$, which is equivalent to $Z$ being on $(AH)$.

Let $f$ denote an inversion centered at $A$ with radius $\sqrt{AN\cdot AB}$. Then $f$ swaps $Z$ and $W$. Unsurprisingly, $f$ also swaps $(AH)$ and $BC$, so we win. $\blacksquare$
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ItsBesi
146 posts
#194
Y by
Nothing new but posting it for storage.

Let $D$ be the feet of the altitude from $A$ to $BC$, let $\omega_1 \cap \omega_2=\{Z\}$.

Claim: Points $\overline{X-Z-Y}$ are collinear.

Proof:

$\angle XZW \stackrel{\omega_1}{=}=90$ and $\angle YZW \stackrel{\omega_1}{=}=90$

Hence $\angle XZY=\angle XZW+\angle YZW=180  \implies \angle XZY=180  \implies$ Points $\overline{X-Z-Y}$ are collinear. $\square$

Claim: Points $\overline{A-Z-W}$ are collinear.

Proof: First note that $\omega_1 \cap \omega_2 =\{W,Z\}$ so $ZW$ is the radical axis of $\omega_1$ and $\omega_2$

Also its well known that points $B,N,M$ and $C$ are concyclic so by Power of the Point Theorem we get:

$Pow(A,\omega_1)=AN \cdot AB=Pow(A, \odot(BNMC))=AM \cdot AC=Pow(A,\omega_2) \implies Pow(A,\omega_1)=Pow(A,\omega_2)$

So $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ hence $A$ lies on the line $WZ \therefore$ Points $\overline{A-Z-W}$ are collinear. $\square$

Claim: Points $W,D,H$ and $Z$ are concyclic.

Proof: It's also well known that points $D,N,B$ and $H$ are concyclic

So by POP we get:
$AH \cdot AD=Pow(A,\odot(DNBH))=AN \cdot AB=Pow(A,\omega_1)=AZ \cdot AW \implies AH \cdot AD= AZ \cdot AW$

which by the converse of POP we have that: Points $W,D,H$ and $Z$ are concyclic. $\square$ Let $\odot(WDHZ)=\Gamma$.

Claim: Points $\overline{X-H-Y}$ are collinear.

Proof:

From $\Gamma \implies \angle HZW \stackrel{\Gamma}{=} 180-\angle HDZ=180-90=90=\angle XZW \implies \angle HZW=\angle XZW \implies$
Points $\overline{X-H-Z}$ are collinear, combining with the first claim that Points $\overline{X-Z-Y}$ are collinear we get that $\overline{X-H-Z-Y}$ are all collinear

Hence points $\overline{X-H-Y}$ are collinear $\blacksquare$
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Maximilian113
575 posts
#195
Y by
Let $P$ be the second intersection of $w_1, w_2.$ Note that by Miquel's Theorem, $AMPN$ is cyclic, but clearly $H$ also lies on this circle so $AMPHN$ is cyclic. Therefore, $$\angle NPH = \angle NAH = 90^\circ - \angle ABC = \angle XBN = \angle XPN,$$so $X, H, P$ are collinear. Similarly, $H, P, Y$ are collinear as well and we are done. QED
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Ilikeminecraft
658 posts
#196
Y by
let $P$ be second intersection of $(BNW), (MWC).$
By Radax on those two circles and $(BNMC),$ we get $A,P,W$ are collinear.
furthermore, miquel theorem on $ABC$ with $M, P, W$ as the points, we get $ANHPM$ is cyclic. Hence, $\angle APH = 90.$
However, $\angle BHP = 90 = \angle HPW$ so $X, H, P$ are collinear.
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endless_abyss
52 posts
#197
Y by
Let $P$ denote the second intersection of the circumcircles $B N W$ and $C M W$,

Claim we claim that $X - P - H$ and $Y - P - H$ are collinear

Let $Q$ denote the foot of $A$ onto $B C$, then we know that -

$\angle B H Q = C$
$\angle B H X  = \angle P W B - C$
$\angle Q H P = \angle 180 - \angle P W B$

So, $X - P - H$ are collinear, similarly, $Y - P - H$ are collinear.
$\square$
:starwars:
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bjump
1033 posts
#198
Y by
Let $D$ be the second interssection of $\omega_1$, and $\omega_2$. Since $\angle XDW = \angle YDW = 90^\circ$, $X$, $D$, and $Y$ are collinear. Let $E$ be the foot from $A$, since $AM \cdot AC = AN \cdot AB$. $A$ lies on $WD$ by radical axis. Converse of Radical axis with lines $NB$, $MC$ and $HE$, and $\omega_1$, and $\omega_2$ gives $HEWD$ cyclic, which means that $\angle WDH = 90^\circ$. So the points are collinear.
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joshualiu315
2534 posts
#199
Y by
not necessarily the fastest solution + i took a ridiculous amount of time to find the miquel point


Let $P$ be the second intersection of $\omega_1$ and $\omega_2$. Since $\angle WPX = \angle WPY$, we have that $P$ lies on $\overline{XY}$. Thus, it suffices to show that $X, H, P$ are collinear since showing that $Y,H,P$ are collinear is analogous.

Note that $P$ is the Miquel point of $(WBN)$, $(WCM)$ and $(AMN)$. Then, note that

\[\angle NPX = \angle NWX = 90^\circ - \angle NXW = 90^\circ - \angle NBW = \angle NAH = \angle NPH,\]
so we are done. $\blacksquare$
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Giant_PT
46 posts
#200
Y by
First IMO problem that I solved within 10 minutes :P
Let $\omega_1$ and $\omega_2$ intersect at $Z$ other than $W$, then clearly $X$, $Z$, and $Y$ are collinear since $\measuredangle XZW + \measuredangle WZY = 90^\circ + 90^\circ=0^\circ$. This also shows that line $XZY$ is a line passing through $Z$ that is perpendicular to line $AW$. After this, since $BNMC$ is cyclic, we see that $A$ is the radical center of circles $\omega_1$, $\omega_2$, and $(BNMC)$, which shows that points $A$, $Z$, and $W$ are collinear. Then, using Miquel's theorem yields that $ANZM$ is cyclic, and since we know that $(ANHM)$ is cyclic as well, we get that $ANHZM$ is cyclic. We can clearly see that $HZ$ is a line passing through $Z$ that is perpendicular to line $AW$ since $\measuredangle AZH = \measuredangle ANH = 90^\circ$, which is in fact same as line $XY$ as shown earlier. This finishes the problem $\square$.
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CrazyInMath
460 posts
#201
Y by
chase the wrong angle, waste 20 minutes.
in case of any config issue we have $AB>AC$.

solution
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