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

(→Solution) |
(→Solution) |
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Line 10: | Line 10: | ||

draw((2,2)--(1,1)); | draw((2,2)--(1,1)); | ||

draw((0,0)--(4,2)); | draw((0,0)--(4,2)); | ||

+ | draw((0,2)--(1,1)); | ||

draw(circle((0,1),1)); | draw(circle((0,1),1)); | ||

draw(circle((4,-3),5)); | draw(circle((4,-3),5)); | ||

Line 20: | Line 21: | ||

dot((0,1)); | dot((0,1)); | ||

label("A",(0,0),NW); | label("A",(0,0),NW); | ||

− | label("B",(1,1), | + | label("B",(1,1),SE); |

label("M",(0,2),N); | label("M",(0,2),N); | ||

label("N",(4,2),N); | label("N",(4,2),N); | ||

Line 29: | Line 30: | ||

label("$t$",(7.5,2),N); | label("$t$",(7.5,2),N); | ||

label("P",(2,2),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> | </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>. | 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>. | + | 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>. |

## Revision as of 22:02, 22 December 2020

## Section 2

Let the circles and intersect at two points and , and let be a common tangent of and that touches and at and respectively. If and , evaluate the angle .

## Solution

Let and be the centers of circles and respectively. Also let be the intersection of and line .

Note that is perpendicular to since is a tangent of . In order for to be perpendicular to , must be the point diametrically opposite . Note that is a right angle since it inscribes a diameter. By AA similarity, . This gives that .

By Power of a Point on point with respect to circle , we have that . Using Power of a Point on point with respect to circle gives that . Therefore and . Since , . We now see that is a triangle. Since it is similar to , $\angle PMB \cong \boxed {\angle NMB \cong 45 \degree \cong \frac{\pi}{4}}$ (Error compiling LaTeX. ! Undefined control sequence.).