Difference between revisions of "2010 IMO Problems/Problem 4"
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This implies that the tangent at <math>M</math> is parallel to <math>LK</math> and therefore that <math>M</math> is the midpoint of arc <math>LK</math>. Hence <math>MK=ML</math>. | This implies that the tangent at <math>M</math> is parallel to <math>LK</math> and therefore that <math>M</math> is the midpoint of arc <math>LK</math>. Hence <math>MK=ML</math>. | ||
+ | |||
+ | <asy> | ||
+ | import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=4.48,ymin=-1.99,ymax=3; | ||
+ | pen ccqqqq=rgb(0,1,0), qqzzqq=rgb(0,0,1), evevff=rgb(0.9,0.9,1); | ||
+ | draw((2,2.55)--(4,0),linewidth(1.3)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.3)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.3)); draw((0.76,-1.16)--(3.43,1.97)); draw((3.43,1.97)--(0.08,1.23)); draw((0.08,1.23)--(0.76,-1.16)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.3)+qqzzqq); draw((0,0)--(4,0),linewidth(1.3)+qqzzqq); draw((2,2.55)--(0.76,-1.16)); draw((0,0)--(3.43,1.97)); draw((-1.85,0.81)--(xmax,-0.75*xmax-0.58)); draw((2,2.55)--(-4.16,2.55),linetype("0 4")); draw((0.76,-1.16)--(-1.85,0.81)); draw((-1.85,0.81)--(-4.16,2.55),linetype("0 4")); draw((3.43,1.97)--(-1.85,0.81)); | ||
+ | dot((0,0),ds); label("$K$",(-0.30,0.06),NE*lsf); dot((4,0),ds); label("$L$",(4.2,0.09),NE*lsf); dot((2,2.55),ds); label("$M$",(2.05,2.62),NE*lsf); dot((1.42,0.81),ds); label("$P$",(1.27,0.96),NE*lsf); dot((3.43,1.97),ds); label("$A$",(3.30,2.17),NE*lsf); dot((0.76,-1.16),ds); label("$C$",(0.91,-1.19),NE*lsf); dot((0.08,1.23),ds); label("$B$",(-0.11,1.49),NE*lsf); dot((-1.85,0.81),ds); label("$S$",(-1.8,0.89),NE*lsf); dot((-4.16,2.55),ds); label("$X$",(-4.16, 2.65),NE*lsf); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy> | ||
== See also == | == See also == |
Revision as of 10:30, 16 July 2010
Contents
[hide]Problem
Let be a point interior to triangle (with ). The lines , and meet again its circumcircle at , , respectively . The tangent line at to meets the line at . Show that from follows .
Solution
Solution 1
Without loss of generality, suppose that . By Power of a Point, , so is tangent to the circumcircle of . Thus, . It follows that after some angle-chasing, so as desired.
Solution 2
Let the tangent at to intersect at . We now have that since and are both isosceles, . This yields that .
Now consider the power of point with respect to .
Hence by AA similarity, we have that . Combining this with the arc angle theorem yields that . Hence .
This implies that the tangent at is parallel to and therefore that is the midpoint of arc . Hence .
See also
2010 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |