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Source: 2017 Iran TST second exam day1 p1

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178 posts

bgn

#1 h Apr 23, 2017, 12:38 PM • 2 Y

Y by Adventure10, sami1618

$ABCD$ is a trapezoid with $AB \parallel CD$ . The diagonals intersect at $P$ . Let $\omega _1$ be a circle passing through $B$ and tangent to $AC$ at $A$ . Let $\omega _2$ be a circle passing through $C$ and tangent to $BD$ at $D$ . $\omega _3$ is the circumcircle of triangle $BPC$ .
Prove that the common chord of circles $\omega _1,\omega _3$ and the common chord of circles $\omega _2, \omega _3$ intersect each other on $AD$ .

Proposed by Kasra Ahmadi

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bgn

#2 h Apr 23, 2017, 1:56 PM • 2 Y

Y by Adventure10, sami1618

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PROF65

#3 h Apr 23, 2017, 10:51 PM • 4 Y

Y by rashah76, Adventure10, Mango247, sami1618

lemma : let $MNPQ$ a trapezoid s.t $MN\parallel PQ$ and $MP\cap NQ=R$ .Let $K$ point of the circle $(NPR)$ then the circle $(MKN)$ is tangent to $MP$ at $M$ iff the line $NK$ bisects $MQ$ .
proof : let $L$ the point of intersection of $NK$ with $PQ$ ; $PM$ tangent to $(MNK)$
$\iff \angle KMP=\angle MNK = \angle NLP=\angle KLP\iff LKMP$ is cyclic $\iff \angle MLK=\angle MPK\iff \angle MLN=\angle KPR =\angle KNR=\angle QNL$ since $MN\parallel PQ$ it's equivalent to $MNQL$ is parallelogram which leads to the result .
back to the problem :
applying the lemma we deduce that the chords cut the side $AD$ at its midpoints
RH HAS
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#4 h Aug 28, 2018, 4:11 PM • 7 Y

Y by Digonta1729, potentialenergy, DonaldJ.Trump, Adventure10, Mango247, sami1618, ehuseyinyigit

only lemma proving TST

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1079 posts

#6 h Dec 15, 2020, 9:46 PM • 2 Y

Y by sami1618, ehuseyinyigit

Linearity of a power of a point for the win!! haha

Define $P(X,\omega_1,\omega_2)=P(X,\omega_1)-P(X,\omega_2)$ . Thus, $P(X,\omega_1,\omega_2)$ is linear.
$PC\cdot P(A,\omega_1,\omega_2)+AP\cdot P(C,\omega_1,\omega_2)=AC\cdot P(P,\omega_1,\omega_2)$ $PD\cdot P(B,\omega_1,\omega_2)+PB\cdot P(D,\omega_1,\omega_2)=BD\cdot P(P,\omega_1,\omega_2)$ Notice that $P(B,\omega_1,\omega_2)=-BD^2$ and $P(C,\omega_1,\omega_2)=AC^2$ and $P(P,\omega_1,\omega_2)=PA^2-PD^2$ , therefore,

$PC\cdot P(A,\omega_1,\omega_2)=AC\cdot (PA^2-PD^2)-AP\cdot AC^2$ $PB\cdot P(D,\omega_1,\omega_2)=BD\cdot (PA^2-PD^2)+PD\cdot BD^2$ Let $M$ be the midpoint of $AD$ . We have
$P(M,\omega_1,\omega_2)=\frac{1}{2} (P(A,\omega_1,\omega_2)+P(D,\omega_1,\omega_2))$ I claim this equality:
$\frac{AC\cdot (PA^2-PD^2)-AP\cdot AC^2}{PC}=-\frac{BD\cdot (PA^2-PD^2)+PD\cdot BD^2}{PB}$ $\frac{AC\cdot PD^2+AP\cdot AC\cdot PC}{PC}=\frac{AC\cdot PD^2+AP\cdot AC^2-AC\cdot PA^2}{PC}=\frac{-AC\cdot (PA^2-PD^2)+AP\cdot AC^2}{PC}=\frac{BD\cdot (PA^2-PD^2)+PD\cdot BD^2}{PB}=\frac{BD\cdot PA^2+PD\cdot BD\cdot PB}{PB}$ $PB\cdot (AC\cdot PD^2+AP\cdot AC\cdot PC)=PC\cdot (BD\cdot PA^2+PD\cdot BD\cdot PB)$ $PB\cdot AC\cdot PD^2+PB\cdot AP\cdot AC\cdot PC=PC\cdot BD\cdot PA^2+PC\cdot PD\cdot BD\cdot PB$ $PB\cdot AC\cdot PD^2-PC\cdot PD\cdot BD\cdot PB+PB\cdot AP\cdot AC\cdot PC-PC\cdot BD\cdot PA^2=0$ Since $\frac{PA}{PC}=\frac{PB}{PD}$ , we have that
$PB\cdot PD\cdot (AC\cdot PD-BD\cdot PC)+AP\cdot PC(AC\cdot PB-PA\cdot BD)=0$ $PB\cdot PD\cdot (AC\cdot PD-(PB+PD)\cdot PC)+AP\cdot PC(AC\cdot PB-PA\cdot (PB+PD))=0$ $PB\cdot PD\cdot (AC\cdot PD-PB\cdot PC-PD\cdot PC)+AP\cdot PC(AC\cdot PB-PA\cdot PB-PA\cdot PD)=0$ $PB\cdot PD\cdot (AP\cdot PD-PB\cdot PC)+AP\cdot PC(CP\cdot PB-PA\cdot PD)=0$ Hence, we have showed that $M$ lies on the radical axis of $\omega_1$ and $\omega_2$ .

Now, redefine $P(X,\omega_1,\omega_3)=(X,\omega_1)-P(X,\omega_3)$ . $P(X,\omega_1,\omega_3)$ is linear.
$PC\cdot P(A,\omega_1,\omega_3)+AP\cdot P(C,\omega_1,\omega_3)=AC\cdot P(P,\omega_1,\omega_3)$ $PD\cdot P(B,\omega_1,\omega_3)+PB\cdot P(D,\omega_1,\omega_3)=BD\cdot P(P,\omega_1,\omega_3)$ Notice that $P(B,\omega_1,\omega_3)=0$ and $P(C,\omega_1,\omega_3)=AC^2$ and $P(P,\omega_1,\omega_3)=PA^2$ , therefore,

$PC\cdot P(A,\omega_1,\omega_3)=AC\cdot PA^2-AP\cdot AC^2=AC\cdot AP\cdot (-PC)\implies P(A,\omega_1,\omega_3)=-AC\cdot AP$ $PB\cdot P(D,\omega_1,\omega_3)=BD\cdot PA^2\implies P(D,\omega_1,\omega_3)=\frac{BD\cdot PA^2}{PB}$ We have
$P(M,\omega_1,\omega_3)=\frac{1}{2} (P(A,\omega_1,\omega_3)+P(D,\omega_1,\omega_3))$ $BD\cdot PA=PB\cdot AC\Longleftrightarrow (PD+PB)\cdot PA=PB\cdot (PA+PC)\Longleftrightarrow PA\cdot PD=PB\cdot PC$ Hence, $P(M,\omega_1,\omega_3)=0$ , i.e $M$ lies on the radical axis of $\omega_1,\omega_3$ . By radical axis theorem, $M$ also lies on the radical axis of $\omega_2,\omega_3$ . We are done.

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133 posts

demmy

#7 h Dec 24, 2023, 5:53 AM • 1 Y

Y by sami1618

Redacted

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887 posts

#10 h Sep 21, 2024, 7:45 PM

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Let $M$ be the midpoint of $AD$ , which we claim is the desired intersection point. Define the linear function $f(X)=\mathbb{P}(X,w_1)-\mathbb{P}(X,w_3)$ . Then
$\begin{align*} f(A)&=AP\cdot AC\\ f(B)&=0\\ f(C)&=-AC^2 \end{align*}$ By the linearity, $\frac{f(A)-f(B)}{AB}=\frac{f(D)-f(C)}{CD}$ Solving for $f(D)$ gives, $f(D)=AC\cdot AP\left(\frac{CD}{AB}-\frac{AC}{AP}\right)=AC\cdot AP\left(\frac{CD}{AB}-\frac{CP}{AP}-1\right)=-AC\cdot AP$ Thus $f(M)=0$ so $M$ lies on the common chord of $w_1$ and $w_3$ . By symmetry, we are finished.

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#11 h Nov 12, 2024, 11:35 AM

Y by

totally same with above.
Let $f(Q)=\mathbb{P}(Q,\omega_1)-\mathbb{P}(Q,\omega_3)$ . Let $DB\cap \omega_1$ at $T$ .
$f(A)=-AP \cdot AC$ , $f(B)=0$ , $f(C)=AC^2$ . We have to compute $f(D)$ . From angle chasing $ATCD$ is cyclic. $f(D)=(DT*DB) - (DP * DB)$ $\implies$ $f(D)=DB *TP=AP *AC$ . Let radical axis of $\omega_1$ and $\omega_3$ intersects $AD$ at $W$ . So, $f(A)\frac{WD}{AD}+f(D)\frac{AW}{AD}=f(W)=0$ . Substuting gives us $WD=AW$ which implies desired point is the midpoint of $AD$ . Doing same thing to $\omega_2$ and $\omega_3$ gives us desired result.

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