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 = \angle ABP</math>. It follows that after some angle-chasing, < | + | [[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{aligned} |
− | \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{aligned}</math></center> so <math>ML = MK</math> 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; | + | <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); | ||
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); | 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); | 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); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
− | </asy> | + | </asy></center> |
=== Solution 2 === | === Solution 2 === | ||
− | == See | + | 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>. |
+ | |||
+ | Now consider the power of point <math>S</math> with respect to <math>\Gamma</math>. | ||
+ | |||
+ | <cmath>SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}</cmath> | ||
+ | |||
+ | |||
+ | Hence by AA similarity, we have that <math>\triangle{SPA} \sim \triangle{SBP}</math>. Combining this with the arc angle theorem yields that <math>\angle{SPA}=\angle{SBP}=\angle{PKL}</math>. Hence <math>PS \| LK</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> | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | |||
+ | Since <math>SP = SC</math>, we can construct circle <math>\omega</math> tangent to <math>SP</math> and <math>SC</math> at <math>P</math> and <math>C</math> respectively. Then, we see that the circumcircle of <math>\triangle ABP</math> must be tangent to <math>SP</math> at <math>P</math> by the converse of the radical axis theorem, since <math>\omega</math> is internally tangent to <math>\Gamma</math> at <math>C</math>, and <math>(ABP)</math> meets <math>\Gamma</math> at <math>A</math> and <math>B</math>. Thus, <math>\angle SPB = \angle BAK = \angle BLK</math>, so lines <math>SP</math> and <math>KL</math> are parallel. | ||
+ | |||
+ | It then follows that <math>\angle SCP = \angle SPC = \angle KDC = \angle KLC + \angle LCD</math>. Since <math>SC</math> is tangent to <math>\Gamma</math>, we see that <math>\angle KLC = \angle SCK</math>, and thus <math>\angle KCD = \angle LCD = \angle SCP - \angle KLC</math>. Thus since <math>CM</math> bisects <math>\angle KCL</math> in cyclic quad <math>CKML</math>, we have <math>MK = ML</math>. <math>\blacksquare</math> | ||
+ | |||
+ | == See Also == | ||
{{IMO box|year=2010|num-b=3|num-a=5}} | {{IMO box|year=2010|num-b=3|num-a=5}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 18:32, 2 April 2021
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,
![$\begin{aligned} \widehat{ML} &= \widehat{MA} + 2\angle AKL \\ &= \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL \\ &= 2\left(\angle CPK + \angle KPS\right) - \widehat{KC} \\ &= 2\angle PCS - \widehat{KC} \\ &= \widehat{MK}, \end{aligned}$](http://latex.artofproblemsolving.com/2/4/b/24b44be8959a647cc8cceaadeedda7116b3b8f06.png)
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
.
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
.
Solution 3
Since , we can construct circle
tangent to
and
at
and
respectively. Then, we see that the circumcircle of
must be tangent to
at
by the converse of the radical axis theorem, since
is internally tangent to
at
, and
meets
at
and
. Thus,
, so lines
and
are parallel.
It then follows that . Since
is tangent to
, we see that
, and thus
. Thus since
bisects
in cyclic quad
, we have
.
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 |