NEPAL TST 2025 DAY 2

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#1 h Apr 12, 2025, 8:40 AM

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Consider an acute triangle $\Delta ABC$ . Let $D$ and $E$ be the feet of the altitudes from $A$ to $BC$ and from $B$ to $AC$ respectively.

Define $D_1$ and $D_2$ as the reflections of $D$ across lines $AB$ and $AC$ , respectively. Let $\Gamma$ be the circumcircle of $\Delta AD_1D_2$ . Denote by $P$ the second intersection of line $D_1B$ with $\Gamma$ , and by $Q$ the intersection of ray $EB$ with $\Gamma$ .

If $O$ is the circumcenter of $\Delta ABC$ , prove that $O$ , $D$ , and $Q$ are collinear if and only if quadrilateral $BCQP$ can be inscribed within a circle.

$\textbf{Proposed by Kritesh Dhakal, Nepal.}$

This post has been edited 1 time. Last edited by Tony_stark0094, Apr 13, 2025, 12:37 AM
Reason: typo

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#2 h Apr 12, 2025, 1:31 PM • 2 Y

Y by khan.academy, alexanderhamilton124

My Problem :blush:

:blush:

. We present two solutions.

Solution 1: Let $X$ be the intersection of ray $OD$ and $(BCP)$ , and $H'$ be the reflection of the orthocenter of $\triangle ABC$ across $BC$ .

Notice that,

$\measuredangle BPC = \measuredangle D_1PD_2 = -\measuredangle D_1AD_2 = -2 \measuredangle BAC = -\measuredangle BOC.$
So, $BOCP$ is cyclic. Furthermore, by Power of a Point

$XD \cdot DO = BD \cdot DC = AD \cdot DH'$
$XAOH'$ is cyclic. Also, $P-C-D_2$ is collinear as $\measuredangle AD_2C = 90^{\circ} = \measuredangle AD_2P$ .

Now, $X$ also lies on $(AD_1D_2)$ as

$\begin{align*} \angle AXP &= \angle AXO + \angle OXP \\ &= \angle AH'O + \angle OCD_2 \\ &= \angle H'AA' + \angle OCA + C \\ &= \angle H'AA' + 90 - B + C \\ \end{align*}$
Where $A'$ is the antipode of $A$ . However, we know

$\angle H'AA' = 90 - \angle HA'A = \angle ABH' - 90 = B + \angle HBC - 90 = B - C.$
Therefore, $\angle AXP = 90^{\circ}.$ Hence, $X$ lies in circle with diameter $AP$ which is precisely $(AD_1D_2)$

Now, if we orient $\triangle{ABC}$ such that $X \equiv Q$ then we get the desired result.

Solution 2: As in the solution above, we show that $BOCP$ is cyclic, and $P-C-D_2$ collinear. Let $X = (BOCP) \cap (D_1AD_2).$ We will show that the points $O-D-X$ are collinear. Notice that it suffices to show that $XD$ is the bisector of $\measuredangle BXC$ . The main claim is as follows.

Claim : $\triangle D_1BX \sim \triangle D_2CX.$

Proof: Notice that

$\measuredangle XD_1B = \measuredangle XD_1P = \measuredangle XD_2P = \measuredangle XD_2C.$
Furthermore,

$\measuredangle D_1BX = - \measuredangle XBP = - \measuredangle XCP = \measuredangle XCD_2.$
Hence, we get the desired similarity.

Now, this similarity gives us

$\frac{XB}{D_1B} = \frac{XC}{D_2C} \iff \frac{XB}{DB} = \frac{XC}{DC}.$
Thus, by angle bisector theorem, $XD$ is the angle bisector of $\measuredangle BXC$ , which tells us that $O-D-X$ is collinear. Therefore, again moving the points $A,B,C$ such that $X \equiv Q$ , we win.

Remark: Clearly, in both solutions, $Q$ and $E$ are just chilling in the corner, contributing absolutely nothing. Only $Q$ shows up at the end. We do a little trolling. :rotfl:

:rotfl:

This post has been edited 2 times. Last edited by Thapakazi, Apr 15, 2025, 3:06 AM

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#3 h Apr 12, 2025, 5:21 PM

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AH i suck at geo, took me way too long to do this, that too with ggb.
Claim 1 : $(AD_1D_2P)$ = $(AP)$ .
Pf: $\measuredangle PD_1A = \measuredangle BD_1A = \measuredangle ADB = \frac{\pi}{2}$

Claim 2 : $P - C - D_2$
Pf: $\measuredangle PD_2A= \frac{\pi}{2} = \measuredangle ADC = \measuredangle CD_2A$

Claim 3 : $(BOCP)$
Pf: $\measuredangle BPC = \measuredangle D_1PD_2 = \measuredangle D_1AD_2 = \measuredangle D_1AD + \measuredangle DAD_2 = 2\measuredangle BAD + 2 \measuredangle DAC = 2 \measuredangle BAC = \measuredangle BOC$

Claim 4 : $O \in AP$
Pf: It suffices to show $(ABC)$ tangent to $(AP)$ . Let $H_1$ and $H_2$ be reflection $H$ in $AB$ and $AC$ where $H$ is orthocenter, its well known that $(ABC) = (AH_1H_2)$ . Consider a dilation at $A$ that takes $H$ to $D$ . It takes $H_1$ to $D_1$ and $H_2$ to $D_2$ so it takes $(ABC) = (AH_1H_2)$ to $(AD_1D_2) = (AP)$ so the circles are tangent at $A$ .

Define $S = (BOCP) \cap (AD_1D_2P)$ with $S \neq P$ . Note that $S-O-D$ implies the problem statement.

Invert about $(ABC)$ . Note: $(BOCSP) = (BOC)\rightarrow BC$ $P \rightarrow AO \cap BC = K$ $A-O-P \implies (AD_1D_2PS) = (AP) \rightarrow (AK) = (ADK)$ $\{S, P\} = (BOCSP) \cap (AD_1D_2PS) \rightarrow BC \cap (ADK) = \{D, K\}$
So $S \rightarrow D \implies S-O-D$ as required

This post has been edited 3 times. Last edited by ThatApollo777, Apr 13, 2025, 4:21 AM
Reason: Typo fixed

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#4 h Apr 12, 2025, 6:27 PM

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Sorry if i confused anyone, i forgot to write an important claim, its fixed now

This post has been edited 1 time. Last edited by ThatApollo777, Apr 12, 2025, 6:35 PM
Reason: time

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#5 h Apr 13, 2025, 1:39 AM

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Congrats Thapakazi!
solution
remark

This post has been edited 4 times. Last edited by InterLoop, Apr 13, 2025, 1:43 AM

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#6 h Apr 13, 2025, 2:55 AM

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i'm finally back at geo after a long long time, and eventually i found this simple nice problem and decided to upload my sol here

Assume that $BCQP$ is cyclic $\Longrightarrow Q=(BCP)\cap (AD_1D_2)$ . The case when $\overline{O,D,Q}$ can be easily dealt with. We define $\varphi$ as the inversion w.r.t. $(ABC)$ . It's easy to prove that $\overline{A,O,P}$ and $\overline{D_2,C,P}$ . So we have $\angle{BPC}=\angle{D_1PD_2}=180^\circ-\angle{D_1AD_2}=180^\circ-\angle{BOC}\Longrightarrow$ $BOCP$ is cyclic. Let $P',D_1',D_2'$ be the inverses of $P,D_1,D_2$ w.r.t. $(ABC)$ , respectively. Hence $\varphi:P=(BOC)\cap AO\longleftrightarrow BC\cap AO=P'\Longrightarrow P'\in BC$ . Clearly $\varphi:A\leftrightarrow A$ . Since inversion preserves angles, we have $\angle{AD_1'P'}=180^\circ-\angle{AD_1P}=90^\circ$ . So then $\angle{AD_2'P'}=180^\circ-\angle{AD_2P}=90^\circ$ . Since $\angle{ADP'}=90^\circ$ , we have $AD_1'DP'D_2'$ is cyclic, so in other words, $D=(AD_1'D_2')\cap BC$ . Finally $\varphi:D=(AD_1'D_2')\cap BC\leftrightarrow (AD_1D_2)\cap(BOC)=Q$ . So $D,Q$ are inverses each other, meaning that $\overline{O,D,Q}$ . We're done. $\blacksquare$

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#7 h Apr 14, 2025, 4:09 AM

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There's nothing special with $B$ altitude ,in fact we prove the general statement for $Q \in \Gamma$ .
Sol:- Since $D_1,D_2$ are reflections of $D$ across $AB,AC$ ,we have $AD_1=AD_2$ . Since $AD,AO$ isogonal lines we observe $\measuredangle D_1AO=\measuredangle BAC=\measuredangle OAD_2$ , Thus $AO$ is the common line of symmetry for $\Gamma , (ABC)$ , meaning that they are tangent at $A$ .Since $\angle AD_1B=90^\circ$ , $P$ must be antipode of $A$ in $\Gamma$ . Now , $\measuredangle OPB=\measuredangle APD_1=\measuredangle AD_2D_1=\measuredangle OCB$ ( where the last equality holds since $\Delta AD_1D_2 \sim \Delta OBC$ ) implies $BOCP$ cyclic. Let $AO \cap BC=F$ , By shooting lemma $OF \cdot OP =OB^2$ . Thus if we invert with respect to $(ABC)$ , $A \leftrightarrow A$ , $F \leftrightarrow P$ , $(AF) \leftrightarrow (AP)$ , (the last notation $(AF),(AP)$ represents circle with diameters $AF,AP$ respectively). If $O-D-Q$ are collinear ,then since $D \in (AF)$ and $Q \in (AP)$ (i.e. inverted image of $(AF)$ ) it follows that $P,Q$ are inverses of each other under this inversion , thus $OD \cdot OQ =OB^2$ ,thus by converse of shooting lemma $(BOCPQ)$ is cyclic . Conversely if $(BOCPQ)$ is cyclic then $Q= (BOCP) \cap \Gamma \neq P$ , Thus the inverse of $Q$ under this inversion lets say $Q*= BC \cap (AF) \neq F$ ,Thus $Q*=D$ , collinearity of $O-D-Q$ follows.

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#8 h Apr 14, 2025, 7:58 PM • 1 Y

Y by GeoKing

The idea is to troll you from realizing it's a general problem, from which you do config geo strats to solve

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cursed_tangent1434

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#9 h Apr 16, 2025, 6:20 AM • 1 Y

Y by brute12

The point $E$ has actually no importance except dealing with a minor configuration issue at the end. Nice result! We start off with the following observation.

Claim : Lines $\overline{BD_1}$ , $\overline{CD_2}$ and $\overline{AO}$ concur at $P$ .

Proof : Note that letting $P' = BD_1 \cap CD_2$ ,
$\begin{align*} \measuredangle D_1P'D_2 &= \measuredangle D_1DD_2 + \measuredangle DD_2P' + \measuredangle P'D_1D\\ &= \measuredangle BAC + \measuredangle CDD_2 + \measuredangle D_1DB \\ &= 2\measuredangle BAC \\ &= \measuredangle D_1AD_2 \end{align*}$ which implies that $P'=BD_1 \cap CD_2$ lies on $\Gamma$ and hence $P' \equiv P$ . Further,
$\measuredangle D_1AP = \measuredangle BAC = \measuredangle ABC + \measuredangle BCA = (90 + \measuredangle ABC) + (90 + \measuredangle BCA) = \measuredangle D_1AB + \measuredangle BAO = \measuredangle D_1AO$ which implies that points $A$ , $O$ and $P$ are collinear, finishing the proof of the claim.

Next note that,
$\measuredangle OPC = \measuredangle APD_2 = \measuredangle AD_1D_2 = 90 + \measuredangle BAC = \measuredangle OBC$ from which we have that $OBPC$ is cyclic. Now comes the key claim.

Claim : Let $X = (BOC) \cap \Gamma \ne P$ . Then points $X$ , $O$ and $D$ are collinear.

Proof : Since $OB=OC$ , $O$ is the minor $BC$ arc midpoint in $(BOC)$ . Thus, $\overline{XO}$ is the internal $\angle BXC-$ bisector. Now, let $D' = XO \cap BC$ . Note,
$\measuredangle BD_1X = \measuredangle PD_1X = \measuredangle PAX = \measuredangle PAX$ and
$\measuredangle XBD_1 = \measuredangle XBP = \measuredangle XOP = \measuredangle XOA$ which together imply that $\triangle XBD_1 \sim \triangle XOA$ . A similar argument shows that $\triangle XCD_2 \sim \triangle XOA$ . Thus,
$\frac{XB}{XO} = \frac{BD_1}{OA} \text{ and } \frac{XC}{XO} = \frac{CD_2}{OA}$ whose quotient yeilds,
$\frac{XB}{XC} = \frac{BD_1}{CD_2} = \frac{BD}{CD}$ since $BD_1=BD$ and $CD_2=CD$ due to relfections.

Now, the Angle Bisector Theorem tells us that,
$\frac{BD'}{CD'} = \frac{XB}{XC} = \frac{BD}{CD}$ which is enough to imply $D' \equiv D$ as desired.

Now, if $BCPQ$ is concyclic, $Q$ must be the second intersection of circles $(BOC)$ and $\Gamma$ besides $P$ which implies $O$ , $D$ and $Q$ are collinear by the claim.

If $O$ , $D$ and $Q$ are collinear, $Q$ is one of the intersections of $\overline{OD}$ with $\Gamma$ , one of which is the second intersection of circles $(BOC)$ and $\Gamma$ . However, since $Q$ is the intersection of ray $EB$ with $\Gamma$ it is on the same side of line $BC$ as $Q$ which implies that it must be the desired intersection point.

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#10 h Apr 17, 2025, 9:32 PM • 1 Y

Y by Gato_combinatorio

Inversion Bash! :-D

:-D

Let $A_1$ be the antipode of $A$ with respect to $(ABC)$ , and let $H$ be the orthocenter of $\triangle ABC$ .

Let $\mathcal{I}$ be the inversion centered at $A$ with radius $\sqrt{AB \cdot AC}$ , composed with reflection across the $A$ -angle bisector. Let $\mathcal{I}'$ be the inversion with respect to $(ABC)$ . For any point $X$ , denote $\mathcal{I}(X) = X'$ and $\mathcal{I}'(X) = X^*$ .

We know that $\mathcal{I}(D) = A_1$ , and since $AD = AD_1$ and $\triangle AD_1D \sim \triangle AA_1D_1'$ , it follows that $D_1'$ is the reflection of $A_1$ across line $AC$ . Similarly, $D_2'$ is the reflection of $A_1$ across $AB$ .

Now observe that $BC$ is the midline of triangle $A_1D_1'D_2'$ , so $D_1'D_2' \parallel BC$ .

Since $D_1$ , $B$ , and $P$ are collinear, then $A$ , $D_1'$ , $C$ , $P'$ are concyclic. As $P'$ lies on $D_1'D_2'$ , we get:
$\angle BAO = \angle BCA_1 = \angle P'D_1'C = \angle P'AC \Rightarrow AP' \perp BC.$
Also, since $P' \in D_1'D_2'$ , the reflection of $P'$ over $BC$ lies on $(ABC)$ , which implies $P' = H$ . Thus $P = H'$ , so $A$ , $O$ , and $P$ are collinear, and $P \in (BOC)$ . Therefore:
$P^* = AO \cap BC.$
Now, since $D \in (AP^*)$ , it follows that $D^* \in AP$ , because $A$ , $O$ , and $P^*$ are collinear. Also $D \in BC$ implies $D^* \in (AD_1D_2) \cap (BOC)$ .

Since $Q = (AD_1D_2) \cap OD$ , we conclude $Q = D^*$ , hence $Q \in (BOCP)$ . $\blacksquare$

This post has been edited 1 time. Last edited by hectorleo123, Apr 17, 2025, 9:32 PM

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