Difference between revisions of "2010 IMO Problems/Problem 4"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | [[Without loss of generality]], suppose that <math>AS > BS</math>. By Power of a Point, <math>SP^2 = SC^2 = SB \cdot SA</math>, so <math>\overline{SP}</math> is tangent to the circumcircle of <math>\triangle ABP</math>. Thus, <math>\angle KPS = 180 - \angle SPA = \widehat{AP}/2 = \angle ABP</math>. It follows that after some angle-chasing, <center><math>\begin{ | + | [[Without loss of generality]], suppose that <math>AS > BS</math>. By Power of a Point, <math>SP^2 = SC^2 = SB \cdot SA</math>, so <math>\overline{SP}</math> is tangent to the circumcircle of <math>\triangle ABP</math>. Thus, <math>\angle KPS = 180 - \angle SPA = \widehat{AP}/2 = \angle ABP</math>. It follows that after some angle-chasing, <center><math>\begin{cases} |
\widehat{ML} &= \widehat{MA} + 2\angle AKL \ &= \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL \ &= 2\left(\angle CPK + \angle KPS\right) - \widehat{KC} \ &= 2\angle PCS - \widehat{KC} \ &= \widehat{MK}, | \widehat{ML} &= \widehat{MA} + 2\angle AKL \ &= \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL \ &= 2\left(\angle CPK + \angle KPS\right) - \widehat{KC} \ &= 2\angle PCS - \widehat{KC} \ &= \widehat{MK}, | ||
− | \end{ | + | \end{cases}</math></center> so <math>ML = MK</math> as desired. |
<center><asy> import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-2.22,xmax=4.48,ymin=-1.99,ymax=3; | <center><asy> import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-2.22,xmax=4.48,ymin=-1.99,ymax=3; | ||
pen ccqqqq=rgb(0.8,0,0), qqzzqq=rgb(0,0.6,0), evevff=rgb(0.9,0.9,1); | pen ccqqqq=rgb(0.8,0,0), qqzzqq=rgb(0,0.6,0), evevff=rgb(0.9,0.9,1); |
Revision as of 23:51, 4 January 2016
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 |