1/sqrt(5) ???

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1/sqrt(5) ???

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Source: Own. Malaysian IMO TST 2025 P12

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#1 h Saturday at 1:10 PM • 1 Y

Y by pingupignu

Two circles $\omega_1$ and $\omega_2$ are externally tangent at a point $A$ . Let $\ell$ be a line tangent to $\omega_1$ at $B\neq A$ and $\omega_2$ at $C\neq A$ . Let $BX$ and $CY$ be diameters in $\omega_1$ and $\omega_2$ respectively. Suppose points $P$ and $Q$ lies on $\omega_2$ such that $XP$ and $XQ$ are tangent to $\omega_2$ , and points $R$ and $S$ lies on $\omega_1$ such that $YR$ and $YS$ are tangent to $\omega_1$ .

a) Prove that the points $P$ , $Q$ , $R$ , $S$ lie on a circle $\Gamma$ .

b) Prove that the four segments $XP$ , $XQ$ , $YR$ , $YS$ determine a quadrilateral with an incircle $\gamma$ , and its radius is $\displaystyle\frac{1}{\sqrt{5}}$ times the radius of $\Gamma$ .

Proposed by Ivan Chan Kai Chin

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#2 h Saturday at 1:38 PM

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Nice problem!

Solved with hints from the proposer himself, Ivan. We assume $B$ , $R$ , $S$ and $C$ , $P$ , $Q$ to appear in that order on $\omega_1$ and $\omega_2$ respectively. Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively. Let $k$ be the tangent to $\omega_1$ and $\omega_2$ passing through $A$ , and let $k$ meet $\ell$ at $Z$ . We see that $ZB^2=ZA^2=ZC^2$ so $\angle BAC = 90^\circ$ , implying $B$ , $A$ , $Y$ and $C$ , $A$ , $X$ are collinear. Define some objects as follows:

$U = YR \cap XP$ ,
$V = YS \cap XQ$ ,
$I = XO_2 \cap YO_1$ ,
$D = IA \cap \ell$ ,
$S_e = \ell \cap O_1O_2 \cap XY$ ,
$\Omega = (S_e, S_eA)$ ,
$K = \Omega \cap BC$ such that $K$ lies on segment $BC$ .

Step 1: $Z = PQ \cap RS$
Proof:
Since $X = PP \cap QQ \in AC$ , $Z = AA \cap CC \in PQ$ by La Hire's Theorem. Similarly $Z \in RS$ . $\square$

Now $ZP \cdot ZQ = ZA^2 = ZR \cdot ZS \implies PQRS \text{ cyclic}.$ This solves a).

Step 2: $XUYV$ has an incircle $\gamma$ .
Proof:
Redefine $\gamma$ such that it is internally tangent to $YR$ , $YS$ and $XQ$ . From Monge's theorem on $\gamma$ , $\omega_1$ and $\omega_2$ we yield $X$ is the exsimilicenter of $\omega_2$ and $\gamma$ . Hence $XP$ is also tangent to $\gamma$ . $\square$

Step 3: $U$ , $V$ $\in k$ .
Proof:
Note that $XO_2$ and $YO_1$ are perpendicular bisectors of $PQ$ and $RS$ , respectively, and are also the respective internal angle bisectors of $\angle UXV$ and $\angle UYV$ , we must have $I = XO_2 \cap YO_1$ is the center of $\gamma$ and $\Gamma$ .
Then, $UI$ must bisect $\angle XUY = \angle RUP$ , combined with $IR=IP$ , we have $UR=UP$ . Since $UR^2=UP^2$ is the power of $U$ to $\omega_1$ and $\omega_2$ , $U$ must lie on their radical axis, namely $k$ . Similarly $V \in k$ . $\square$

Step 4: $P$ , $R$ , $S_e$ collinear.
Proof:
I first claim that $K = YP \cap XR$ . Let $K' = YP \cap \ell$ .

In view of $PP$ , $AC$ , $YY_1$ concur, where $Y_1 = XY \cap \omega_2$ , from the Dual of Desargues Involution Theorem we yield some reciprocal pairs on $\omega_2$ , that is, $\{(P,P), (A,C), (Y,Y_1)\}.$ By perspectivity at $Y$ we have $\{(K',K'), (B,C), (\infty_{\ell}, S_e)\}$ being reciprocal pairs of some involution on $\ell$ . This gives $S_eB \cdot S_eC = S_eK'^2$ while the LHS is $S_eA^2$ . Hence $S_eK'=S_eA$ , from which follows that $K' \equiv K$ . Similarly $K \in XR \implies K = YP \cap XR$ .

Next, we observe that $YP \cdot YK = YC^2 = YA \cdot YB = YR^2$ meaning that $(RPK)$ is tangent to $\omega_1$ . Similarly it must also be tangent to $\omega_2$ . From Monge's theorem we have $S_e \in RP$ . $\square$

Note that similarly, $S_e$ , $Q$ , $S$ collinear.

Step 5: $IA \bot \ell$ .
Proof:
Let $D'$ be the inverse of $D$ in $\Omega$ , so that $AD \bot \ell$ and $S_eD' \cdot S_eZ = S_eA^2 = S_eP \cdot S_eR = S_eQ \cdot S_eS\implies D' = (ZRP) \cap (ZSQ).$ Hence $D'$ is the Miquel point of $RPQS$ $\implies ID' \bot \ell$ . So $D' \equiv D$ . $\square$

Step 6: $A \in \gamma$ .
Proof:
Since $Y$ is the exsimilicenter of $\omega_1$ and $\gamma$ and $IA \parallel O_1B$ , the homothety centered at $Y$ sending $\gamma$ to $\omega_1$ sends $A$ to $B$ , i.e. $A \in \gamma$ . $\square$

Lemma: If $(A, B; X, Y)=-1$ and $X$ lies between $A$ and $B$ , such that $AX=a$ , $BX=b$ , then we have $\frac{XY}{YB} = \frac{2a}{a+b}$ .
Proof:
$\frac{AY}{YB} = \frac{a}{b} \implies 1 + \frac{a+b}{YB} = \frac{a}{b} \implies YB = \frac{b(a+b)}{a-b}$ So $XY = b + \frac{b(a+b)}{a-b} = \frac{2ab}{a-b} \implies \frac{XY}{YB} = \frac{2a}{a+b}. \square$
Step 7: $AD=2IA$ .
Proof:
We have, from $O_1B \parallel IA \parallel O_2C$ , $\frac{IA}{r_2} = \frac{r_1}{r_1+r_2}$ and $\frac{AD}{r_2} = \frac{S_eA}{S_eO_2}$ Applying the lemma gives $\frac{2IA}{r_2} = \frac{AD}{r_2} \implies AD = 2IA$ . $\square$

Let $A_1 = IA \cap \Omega$ . Since $S_eR \cdot S_eP = S_eA^2 \implies \Gamma$ is orthogonal to $\Omega$ , we have $IA \cdot IA_1 = IR^2$ . Now, $AD=DA_1 \implies IA_1 = 5IA \implies 5IA^2 = IR^2.$ Hence the radius of $\gamma$ is $\frac{1}{\sqrt{5}}$ times the radius of $\Gamma$ , completing part b).

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everythingpi3141592

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everythingpi3141592

#3 h Yesterday at 1:20 PM

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Let $O_1$ , $r_1$ and $O_2$ , $r_2$ be the centre and radii of $\omega_1$ and $\omega_2$ respectively

Claim: $XP = XQ = 2R_1$ .

Proof: Note that $X, A, C$ and $Y, A, B$ are collinear due to homotethy. Consider the inversion at $X$ with radius $XB$ . Then, note that $\omega_1$ and the common tangent $BC$ get swapped. So, $\omega_2$ goes to a circle that is tangent to $BC$ at $C$ and $\omega_1$ at $A$ , but this is in fact $\omega_2$ itself, so the circle we described must be orthogonal to $\omega_2$ . This implies the result.

Now, let $XO_2$ and $YO_1$ intersect at $Z$ . Then, $Z$ divides both lines in the ratio $\frac{R_1}{R_2}$ because the homotethy at this centre swaps $XO_1$ and $YO_2$ . Now, what this means is that if the foot from $Z$ onto $XP$ is $P'$ , then $PP' = \frac{2R_1R_2}{R_1+R_2}$ , and $ZP' = \frac{R_1R_2}{R_1+R_2}$ . This, implies the result

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#4 h Yesterday at 2:48 PM

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Cool Problem

Claim 1: $PQRS$ is cyclic.
Proof: Let $AA \cap \ell = V$ then it is suffices to prove that $PQ,RS$ pass through $V$ by power of point, by La Hire, it suffices to show that $A,C,X$ are collinear but this is obvious as $\measuredangle BAC = \measuredangle BAX = 90^{\circ}$ $\blacksquare$

Claim 2: $XB=XQ=XP$
Proof: It suffices to prove that the circle centered at $X$ with radius $XB$ is orthogonal to $\omega_2$ , invert at $X$ with radius $XB$ it then suffices to prove that $\omega_2$ is fixed, note that $A,C$ exchange as $XB^2 = XA \cdot XC$ , and note that $P,Q$ remain fixed and so $\omega_2 \rightarrow (ACPQ) = \omega_2$ under inversion so we're done. $\blacksquare$

Claim 3: $XP,XQ,YR,YS$ form a quadrilateral with incircle of radius $\frac{1}{\sqrt{5}}$ th of $(PQRS)$ .
Proof: Let $O_1,O_2$ be the centers of $\omega_1,\omega_2$ , $XP \cap YR = U$ and $XQ \cap YS = V$ , Let $XO_2 \cap YO_1=I$ , I claim that $I$ is the incenter of $XUVY$ , indeed we compute distance from $I$ to $XP$ and proving it symmetric in radiuses of $\omega_1$ ad $\omega_2$ suffices by symmetry, indeed notice that under homethety at $I$ we take $XO_1$ to $O_2Y$ , thus we must have $\text{dist}(I,XP) = \frac{r_1r_2}{r_1+r_2}$ , next we claim that $I$ is center of $PQRS$ , this is obvious as it lies on the perpendicular bisectors of $PQ,RS$ by definition, now to finish we compute the radius of $(PQRS)$ , let the foot of perpendicular from $I$ to $XP$ be $L$ then $LP = \frac{2r_1r_2}{r_1+r_2}$ by using the homothety at $I$ and by pythogoras we are done. $\blacksquare$

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