Difference between revisions of "2010 IMO Problems/Problem 4"
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Let the tangent at <math>M</math> to <math>\Gamma</math> intersect <math>SC</math> at <math>X</math>. We now have that since <math>\triangle{XMC}</math> and <math>\triangle{SPC}</math> are both isosceles, <math>\angle{SPC}=\angle{SCP}=\angle{XMC}</math>. This yields that <math>MX \| PS</math>. | Let the tangent at <math>M</math> to <math>\Gamma</math> intersect <math>SC</math> at <math>X</math>. We now have that since <math>\triangle{XMC}</math> and <math>\triangle{SPC}</math> are both isosceles, <math>\angle{SPC}=\angle{SCP}=\angle{XMC}</math>. This yields that <math>MX \| PS</math>. | ||
Revision as of 23:08, 15 July 2010
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.
![[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); filldraw(arc((0.08,1.23),0.37,-17.48,12.32)--(0.08,1.23)--cycle,evevff,blue); filldraw(arc((0,0),0.37,0,29.8)--(0,0)--cycle,evevff,blue); filldraw(arc((1.42,0.81),0.37,-179.98,-150.2)--(1.42,0.81)--cycle,evevff,blue); draw((2,2.55)--(4,0),linewidth(1.6)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.6)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.2)); 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((3.43,1.97)--(xmin,0.22*xmin+1.22)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.6)+qqzzqq); draw((0,0)--(4,0)); draw(circle((1.42,3.15),2.34),linetype("4 4")); draw(arc((0.08,1.23),0.37,-17.48,12.32),blue); draw(arc((0.08,1.23),0.31,-17.48,12.32),blue); draw(arc((0,0),0.37,0,29.8),blue); draw(arc((0,0),0.31,0,29.8),blue); draw(arc((1.42,0.81),0.37,-179.98,-150.2),blue); draw(arc((1.42,0.81),0.31,-179.98,-150.2),blue); 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((-1.85,0.81)--(0.76,-1.16),linewidth(1.6)+qqzzqq); 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); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/a/b/8/ab8e42b0fbc1f1cbbf674e77438ab801c52c771f.png)
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 .
![\[SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}\]](http://latex.artofproblemsolving.com/0/1/c/01c2b22c06bde3f2954f728fe145ef4476f2bbe6.png)
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 | ||
