PA^2 \cdot QB = QA^2 \cdot PB inside an equilateral, circumcircle rlted

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PA^2 \cdot QB = QA^2 \cdot PB inside an equilateral, circumcircle rlted

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#1 h Jul 10, 2019, 9:30 AM • 1 Y

Y by Adventure10

In the equilateral triangle $ABC, M, N$ are the midpoints of the sides $AB, AC$ , respectively. The line $MN$ intersects the circumcircle of $\vartriangle ABC$ at $K$ and $L$ and the lines $CK$ and $CL$ meet the line $AB$ at $P$ and $Q$ , respectively. Prove that $PA^2 \cdot QB = QA^2 \cdot PB$ .

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#2 h Mar 30, 2023, 4:57 AM

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In fact, we may compute each of the individual lengths.
$[asy] size(8cm); pair C = dir(90); pair B = dir(210); pair A = dir(330); pair M = (A+B)/2; pair N = (A+C)/2; pair K = intersectionpoints(10*M-9*N--10*N-9*M,unitcircle)[0]; pair L = intersectionpoints(10*M-9*N--10*N-9*M,unitcircle)[1]; pair P = extension(C,K,A,B); pair Q = extension(C,L,A,B); pair O = (0,0); pair S = (M+N)/2; draw(unitcircle); draw(B--Q--C--B--cycle); draw(A--C); draw(C--K--L); draw(C--O--S); draw(O--N); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(315)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$N$",N,dir(N)); dot("$P$",P,dir(135)); dot("$O$",O,dir(270)); dot("$Q$",Q,dir(Q)); [/asy]$
Let the radius of the circumcircle be $R=4t$ and the centre be $O$ . Ledt $S$ be the foot of the perpendicular from $O$ to $MN$ . Then
$OS=ON/2=t,\quad KS=\sqrt{16t^2-t^2}=\sqrt{15}t,\quad MS=\sqrt{4t^2-t^2}=\sqrt{3}t.$ Thus,
$KM=KS-MS=(\sqrt{15}-\sqrt{3})t.$ Since $\triangle PMK\sim \triangle PBC$ and $BC=\sqrt{24}t$ , we have
$\frac{PM}{PB}=\frac{MK}{BC}=\frac{\sqrt{5}-1}{4}$ $PM=\frac{\sqrt{5}-1}{4}PB.$ Thus,
$\begin{align*} PA &= PM+MA=PM+BM=2PM+PB=\frac{1}{2}(\sqrt{5}+1)PB \\ BA &= PA+PB=\frac{1}{2}(\sqrt{5}+3)PB. \end{align*}$ Therefore, $PA^2=PB\cdot BA$ . Similarly, $QA^2=QB\cdot BA$ . Thus $PA^2\cdot QB=QA^2\cdot PB$ .

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#3 h Mar 30, 2023, 5:00 AM

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Alternatively:

Note that $\angle BCL=\angle CBK$ and $\angle BKC=60^\circ$ . Thus,
$\begin{align*} &\angle BCL+\angle BCP = \angle CBK+\angle BCK=180^\circ - \angle BKC = 120^\circ \\ \Rightarrow &\angle BQC = 180^\circ - \angle CBQ-\angle BCQ = 120^\circ - \angle BCL = \angle BCP \\ \Rightarrow & \triangle BCQ\sim \triangle BPC \\ \Rightarrow & \frac{CQ}{PC}=\frac{BQ}{BC}=\frac{BC}{BP} \\ \Rightarrow & \left(\frac{CQ}{PC}\right)^2 = \frac{BQ}{BC}\cdot \frac{BC}{BP}=\frac{BQ}{BP}. \end{align*}$ Now since $KL\parallel BC$ , $A$ is also the midpoint along the arc of $KL$ . Thus, $\angle KCA=\angle ACL$ . By the angle bisector theorem on $\triangle PCQ$ , we have
$\frac{CQ}{CP}=\frac{AQ}{AP}.$ Combining, we have
$\left(\frac{AQ}{AP}\right)^2=\left(\frac{CQ}{PC}\right)^2=\frac{BQ}{BP}$ and the conclusion follows.

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