Difference between revisions of "2012 JBMO Problems/Problem 2"

Revision as of 22:02, 22 December 2020

Section 2

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 $\triangle MPA$ , $\angle PMB \cong \boxed {\angle NMB \cong 45 \degree \cong \frac{\pi}{4}}$ (Error compiling LaTeX. Unknown error_msg).

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Difference between revisions of "2012 JBMO Problems/Problem 2"

Revision as of 22:02, 22 December 2020

Section 2

Solution

@@ Line 10: / Line 10: @@
 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));
@@ Line 20: / Line 21: @@
 dot((0,1));
 label("A",(0,0),NW);
-label("B",(1,1),NW);
+label("B",(1,1),SE);
 label("M",(0,2),N);
 label("N",(4,2),N);
@@ Line 29: / Line 30: @@
 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 <math>O_1</math> and <math>O_2</math> be the centers of circles <math>k_1</math> and <math>k_2</math> respectively. Also let <math>P</math> be the intersection of <math>\overrightarrow{AB}</math> and line <math>t</math>.
-Note that <math>\overline{O_1M}</math> is perpendicular to <math>\overline{MN}</math> since <math>M</math> is a tangent of <math>k_1</math>. In order for <math>\overline{AM}</math> to be perpendicular to <math>\overline{MN}</math>, <math>A</math> must be the point diametrically opposite <math>M</math>.   (to be continued.)
+Note that <math>\overline{O_1M}</math> is perpendicular to <math>\overline{MN}</math> since <math>M</math> is a tangent of <math>k_1</math>. In order for <math>\overline{AM}</math> to be perpendicular to <math>\overline{MN}</math>, <math>A</math> must be the point diametrically opposite <math>M</math>. Note that <math>\angle MBA</math> is a right angle since it inscribes a diameter. By AA similarity, <math>\triangle ABM\sim\triangle AMP</math>. This gives that <math>\angle BMA \cong \angle MPA</math>.
+By [[Power of a Point]] on point <math>P</math> with respect to circle <math>k_1</math>, we have that <math>PM^2=PB\cdot PA</math>. Using Power of a Point on point <math>P</math> with respect to circle <math>k_2</math> gives that <math>PN^2=PB\cdot PA</math>. Therefore <math>PM^2=PN^2</math> and <math>PM=PN</math>. Since <math>MN=2AM</math>, <math>MA=MP</math>. We now see that <math>\triangle APM</math> is a <math>45-45-90</math> triangle. Since it is similar to <math>\triangle MPA</math>, <math>\angle PMB \cong \boxed {\angle NMB \cong 45 \degree \cong \frac{\pi}{4}}</math>.