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2010 IMO Problems/Problem 4

Problem

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Solution

Solution 1

Without loss of generality, suppose that $AS > BS$. By Power of a Point, $SP^2 = SC^2 = SB \cdot SA$, so $\overline{SP}$ is tangent to the circumcircle of $\triangle ABP$. Thus, $\angle KPS = 180 - \angle SPA = \widehat{AP}/2 = \angle ABP$. It follows that after some angle-chasing,

$\begin{aligned} \widehat{ML} &= \widehat{MA} + 2\angle AKL \\ &= \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL \\ &= 2\left(\angle CPK + \angle KPS\right) - \widehat{KC} \\ &= 2\angle PCS - \widehat{KC} \\ &= \widehat{MK}, \end{aligned}$

so $ML = MK$ as desired.

[asy] import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-2.22,xmax=4.48,ymin=-1.99,ymax=3; pen ccqqqq=rgb(0.8,0,0), qqzzqq=rgb(0,0.6,0), evevff=rgb(0.9,0.9,1); filldraw(arc((0.08,1.23),0.37,-17.48,12.32)--(0.08,1.23)--cycle,evevff,blue); filldraw(arc((0,0),0.37,0,29.8)--(0,0)--cycle,evevff,blue); filldraw(arc((1.42,0.81),0.37,-179.98,-150.2)--(1.42,0.81)--cycle,evevff,blue); draw((2,2.55)--(4,0),linewidth(1.6)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.6)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.2)); draw((0.76,-1.16)--(3.43,1.97)); draw((3.43,1.97)--(0.08,1.23)); draw((0.08,1.23)--(0.76,-1.16)); draw((3.43,1.97)--(xmin,0.22*xmin+1.22)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.6)+qqzzqq); draw((0,0)--(4,0)); draw(circle((1.42,3.15),2.34),linetype("4 4")); draw(arc((0.08,1.23),0.37,-17.48,12.32),blue); draw(arc((0.08,1.23),0.31,-17.48,12.32),blue); draw(arc((0,0),0.37,0,29.8),blue); draw(arc((0,0),0.31,0,29.8),blue); draw(arc((1.42,0.81),0.37,-179.98,-150.2),blue); draw(arc((1.42,0.81),0.31,-179.98,-150.2),blue); draw((2,2.55)--(0.76,-1.16)); draw((0,0)--(3.43,1.97)); draw((-1.85,0.81)--(xmax,-0.75*xmax-0.58)); draw((-1.85,0.81)--(0.76,-1.16),linewidth(1.6)+qqzzqq); dot((0,0),ds); label("$K$",(-0.30,0.06),NE*lsf); dot((4,0),ds); label("$L$",(4.2,0.09),NE*lsf); dot((2,2.55),ds); label("$M$",(2.05,2.62),NE*lsf); dot((1.42,0.81),ds); label("$P$",(1.27,0.96),NE*lsf); dot((3.43,1.97),ds); label("$A$",(3.30,2.17),NE*lsf); dot((0.76,-1.16),ds); label("$C$",(0.91,-1.19),NE*lsf); dot((0.08,1.23),ds); label("$B$",(-0.11,1.49),NE*lsf); dot((-1.85,0.81),ds); label("$S$",(-1.8,0.89),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Solution 2

Let the tangent at $M$ to $\Gamma$ intersect $SC$ at $X$. We now have that since $\triangle{XMC}$ and $\triangle{SPC}$ are both isosceles, $\angle{SPC}=\angle{SCP}=\angle{XMC}$. This yields that $MX \| PS$.

Now consider the power of point $S$ with respect to $\Gamma$.

\[SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}\]


Hence by AA similarity, we have that $\triangle{SPA} \sim \triangle{SBP}$. Combining this with the arc angle theorem yields that $\angle{SPA}=\angle{SBP}=\angle{PKL}$. Hence $PS \| LK$.

This implies that the tangent at $M$ is parallel to $LK$ and therefore that $M$ is the midpoint of arc $LK$. Hence $MK=ML$.

[asy] import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=4.48,ymin=-1.99,ymax=3; pen ccqqqq=rgb(0,1,0), qqzzqq=rgb(0,0,1), evevff=rgb(0.9,0.9,1); draw((2,2.55)--(4,0),linewidth(1.3)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.3)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.3)); draw((0.76,-1.16)--(3.43,1.97)); draw((3.43,1.97)--(0.08,1.23)); draw((0.08,1.23)--(0.76,-1.16)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.3)+qqzzqq); draw((0,0)--(4,0),linewidth(1.3)+qqzzqq); draw((2,2.55)--(0.76,-1.16)); draw((0,0)--(3.43,1.97)); draw((-1.85,0.81)--(xmax,-0.75*xmax-0.58)); draw((2,2.55)--(-4.16,2.55),linetype("0 4")); draw((0.76,-1.16)--(-1.85,0.81)); draw((-1.85,0.81)--(-4.16,2.55),linetype("0 4")); draw((3.43,1.97)--(-1.85,0.81)); dot((0,0),ds); label("$K$",(-0.30,0.06),NE*lsf); dot((4,0),ds); label("$L$",(4.2,0.09),NE*lsf); dot((2,2.55),ds); label("$M$",(2.05,2.62),NE*lsf); dot((1.42,0.81),ds); label("$P$",(1.27,0.96),NE*lsf); dot((3.43,1.97),ds); label("$A$",(3.30,2.17),NE*lsf); dot((0.76,-1.16),ds); label("$C$",(0.91,-1.19),NE*lsf); dot((0.08,1.23),ds); label("$B$",(-0.11,1.49),NE*lsf); dot((-1.85,0.81),ds); label("$S$",(-1.8,0.89),NE*lsf); dot((-4.16,2.55),ds); label("$X$",(-4.16, 2.65),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]

Solution 3

Since $SP = SC$, we can construct circle $\omega$ tangent to $SP$ and $SC$ at $P$ and $C$ respectively. Then, we see that the circumcircle of $\triangle ABP$ must be tangent to $SP$ at $P$ by the converse of the radical axis theorem, since $\omega$ is internally tangent to $\Gamma$ at $C$, and $(ABP)$ meets $\Gamma$ at $A$ and $B$. Thus, $\angle SPB = \angle BAK = \angle BLK$, so lines $SP$ and $KL$ are parallel.

It then follows that $\angle SCP = \angle SPC = \angle KDC = \angle KLC + \angle LCD$. Since $SC$ is tangent to $\Gamma$, we see that $\angle KLC = \angle SCK$, and thus $\angle KCD = \angle LCD = \angle SCP - \angle KLC$. Thus since $CM$ bisects $\angle KCL$ in cyclic quad $CKML$, we have $MK = ML$. $\blacksquare$

See Also

2010 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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