Difference between revisions of "2012 JBMO Problems/Problem 2"
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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^{\circ} \cong \frac{\pi}{4}}</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^{\circ} \cong \frac{\pi}{4}}</math>. | ||
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+ | Solution by Someonenumber011 :) | ||
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+ | {{JBMO box|year=2012|before=[[2011 JBMO]]|after=[[2013 JBMO]]}} | ||
+ | [[Category: JBMO]] |
Revision as of 22:04, 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 , .
Solution by Someonenumber011 :)
2012 JBMO (Problems • Resources) | ||
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