Difference between revisions of "2012 JBMO Problems/Problem 2"
(→Solution) |
(→Solution) |
||
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. Unknown error_msg).