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

(Solution)
(Solution)
Line 9: Line 9:
 
draw((0,0)--(2,2));
 
draw((0,0)--(2,2));
 
draw((2,2)--(1,1));
 
draw((2,2)--(1,1));
 +
draw((0,0)--(4,2));
 
draw(circle((0,1),1));
 
draw(circle((0,1),1));
 
draw(circle((4,-3),5));
 
draw(circle((4,-3),5));
Line 24: Line 25:
 
label("$O_1$",(0,1),NW);
 
label("$O_1$",(0,1),NW);
 
label("$O_2$",(4,-3),NE);
 
label("$O_2$",(4,-3),NE);
label("$k_1$",(-0.7,1.7),NW);
+
label("$k_1$",(-0.7,1.63),NW);
 
label("$k_2$",(7.6,0.46),NE);
 
label("$k_2$",(7.6,0.46),NE);
 
label("$t$",(7.5,2),N);
 
label("$t$",(7.5,2),N);
Line 30: Line 31:
 
</asy>
 
</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.
+
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.)

Revision as of 22:32, 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(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),NW); 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); [/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 $M$. (to be continued.)

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