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2012 JBMO Problems/Problem 2: Difference between revisions

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Solution by Someonenumber011 :)
Solution by Someonenumber011 :)
The last paragraph is basically using the fact that <math>P</math> lies on the radical axis of <math>k_1</math> and <math>k_2</math>.


{{JBMO box|year=2012|before=[[2011 JBMO]]|after=[[2013 JBMO]]}}
{{JBMO box|year=2012|before=[[2011 JBMO]]|after=[[2013 JBMO]]}}
[[Category: JBMO]]
[[Category: JBMO]]

Revision as of 19:36, 22 August 2024

Problem

Let the circles $k_1$ and $k_2$ intersect at two points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$ that touches $k_1$ and $k_2$ at $M$ and $N$ respectively. If $t\perp AM$ and $MN=2AM$, evaluate the angle $NMB$.

Solution

[asy] size(15cm,0); draw((0,0)--(0,2)--(4,2)--(4,-3)--(0,0)); draw((-1,2)--(9,2)); draw((0,0)--(2,2)); draw((2,2)--(1,1)); draw((0,0)--(4,2)); draw((0,2)--(1,1)); draw(circle((0,1),1)); draw(circle((4,-3),5)); dot((0,0)); dot((0,2)); dot((2,2)); dot((4,2)); dot((4,-3)); dot((1,1)); dot((0,1)); label("A",(0,0),NW); label("B",(1,1),SE); label("M",(0,2),N); label("N",(4,2),N); label("$O_1$",(0,1),NW); label("$O_2$",(4,-3),NE); label("$k_1$",(-0.7,1.63),NW); label("$k_2$",(7.6,0.46),NE); label("$t$",(7.5,2),N); label("P",(2,2),N); draw(rightanglemark((0,0),(0,2),(2,2))); draw(rightanglemark((0,2),(1,1),(2,2))); draw(rightanglemark((0,2),(4,2),(4,0))); [/asy]

Let $O_1$ and $O_2$ be the centers of circles $k_1$ and $k_2$ respectively. Also let $P$ be the intersection of $\overrightarrow{AB}$ and line $t$.

Note that $\overline{O_1M}$ is perpendicular to $\overline{MN}$ since $M$ is a tangent of $k_1$. In order for $\overline{AM}$ to be perpendicular to $\overline{MN}$, $A$ must be the point diametrically opposite $M$. Note that $\angle MBA$ is a right angle since it inscribes a diameter. By AA similarity, $\triangle ABM\sim\triangle AMP$. This gives that $\angle BMA \cong \angle MPA$.

By Power of a Point on point $P$ with respect to circle $k_1$, we have that $PM^2=PB\cdot PA$. Using Power of a Point on point $P$ with respect to circle $k_2$ gives that $PN^2=PB\cdot PA$. Therefore $PM^2=PN^2$ and $PM=PN$. Since $MN=2AM$, $MA=MP$. We now see that $\triangle APM$ is a $45-45-90$ triangle. Since it is similar to $\triangle MPA$, $\angle PMB \cong \boxed {\angle NMB \cong 45^{\circ} \cong \frac{\pi}{4}}$.

Solution by Someonenumber011 :)

The last paragraph is basically using the fact that $P$ lies on the radical axis of $k_1$ and $k_2$.

2012 JBMO (Problems • Resources)
Preceded by
2011 JBMO 		Followed by
2013 JBMO
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