Difference between revisions of "2016 USAJMO Problems/Problem 1"
(Added Solution 1 and moved previous solution to solution 2.) |
|||
Line 5: | Line 5: | ||
Prove that as <math>P</math> varies, the circumcircle of triangle <math>\triangle PI_BI_C</math> passes through a fixed point. | Prove that as <math>P</math> varies, the circumcircle of triangle <math>\triangle PI_BI_C</math> passes through a fixed point. | ||
− | == Solution == | + | ==Solution 1== |
+ | |||
+ | We claim that <math>M</math> (midpoint of arc <math>BC</math>) is the fixed point. | ||
+ | We would like to show that <math>M</math>, <math>P</math>, <math>I_B</math>, <math>I_C</math> are cyclic. | ||
+ | |||
+ | We extend <math>PI_B</math> to intersect <math>\omega</math> again at R. | ||
+ | We extend <math>PI_C</math> to intersect <math>\omega</math> again at S. | ||
+ | |||
+ | We invert around a circle centered at <math>P</math> with radius <math>1</math> (for convenience). | ||
+ | (I will denote X' as the reflection of X for all the points) | ||
+ | The problem then becomes: Prove <math>I_B'</math>, <math>I_C'</math>, and <math>M'</math> are collinear. | ||
+ | |||
+ | Now we look at triangle <math>\triangle P'R'S'</math>. We apply Menelaus (the version where all three points lie outside the triangle). | ||
+ | It suffices to show that | ||
+ | <cmath>\dfrac{P'I_B'}{I_B'R'} \cdot \dfrac{R'M'}{M'S'} \cdot \dfrac{S'I_C'}{I_C'P'} = 1</cmath> | ||
+ | |||
+ | By inversion, we know <math>P'X' = \dfrac{1}{PX}</math> for any point <math>X</math> and <math>X'Y' = \dfrac{XY}{PX \cdot PY}</math> for any points <math>X</math> and <math>Y</math>. | ||
+ | |||
+ | Plugging this into our Menelaus equation we obtain that it suffices to show | ||
+ | <cmath>\dfrac{\dfrac{1}{PI_B}}{\dfrac{RI_B}{PI_B \cdot PR}} \cdot \dfrac{\dfrac{RM}{PR \cdot PM}}{\dfrac{SM}{PS \cdot PM}} \cdot \dfrac{\dfrac{SI_C}{PI_C \cdot PS}}{\dfrac{1}{PI_C}} = 1</cmath> | ||
+ | We cancel out the like terms and rewrite. It suffices to show | ||
+ | <cmath>\dfrac{RM \cdot SI_C}{SM \cdot RI_B} = 1</cmath> | ||
+ | We know that <math>AM</math> is the diameter of <math>\omega</math> because <math>\triangle ABC</math> is isosceles and <math>AM</math> is the angle bisector. We also know <math>\angle RMA = \dfrac{\angle ACB}{2} = \dfrac{\angle ABC}{2} = \angle SMA</math> so <math>R</math> and <math>S</math> are symmetric with respect to <math>AM</math> so <math>RM = SM</math>. | ||
+ | |||
+ | Thus, it suffices to show <math>\dfrac{SI_C}{RI_B} = 1</math>. | ||
+ | This is obvious because <math>RI_B = RA = SA = SI_C</math>. | ||
+ | Therefore we are done. <math>\Box</math> | ||
+ | |||
+ | == Solution 2== | ||
We will use complex numbers, with the circumcircle of <math>\triangle ABC</math> as the unit circle. Let <cmath>A=1, B=w^2,C=\frac{1}{w^2}, P=p^2,</cmath> such that <cmath>I_B=w+wp-p, I_C=p-\frac{p}{w}+\frac{1}{w}.</cmath> We claim that the circumcircle of <math>\triangle PI_BI_C</math> passes through <math>M=-1.</math> This is true iff <cmath>k=\frac{(I_B-M)(I_C-P)}{(I_B-P)(I_C-M)}=\frac{(w+wp-p+1)(p+\frac{1}{w}-\frac{p}{w}-p^2)}{(w+wp-p-p^2)(p+\frac{1}{w}-\frac{p}{w}+1)}=\frac{(wp+1)(1-p)}{(p+1)(w-p)}</cmath> is real. This is true iff <math>k=\overline{k}.</math> We can compute <cmath>\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,</cmath> so we are done. <math>\blacksquare</math> | We will use complex numbers, with the circumcircle of <math>\triangle ABC</math> as the unit circle. Let <cmath>A=1, B=w^2,C=\frac{1}{w^2}, P=p^2,</cmath> such that <cmath>I_B=w+wp-p, I_C=p-\frac{p}{w}+\frac{1}{w}.</cmath> We claim that the circumcircle of <math>\triangle PI_BI_C</math> passes through <math>M=-1.</math> This is true iff <cmath>k=\frac{(I_B-M)(I_C-P)}{(I_B-P)(I_C-M)}=\frac{(w+wp-p+1)(p+\frac{1}{w}-\frac{p}{w}-p^2)}{(w+wp-p-p^2)(p+\frac{1}{w}-\frac{p}{w}+1)}=\frac{(wp+1)(1-p)}{(p+1)(w-p)}</cmath> is real. This is true iff <math>k=\overline{k}.</math> We can compute <cmath>\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,</cmath> so we are done. <math>\blacksquare</math> |
Revision as of 19:46, 24 June 2016
Contents
Problem
The isosceles triangle , with , is inscribed in the circle . Let be a variable point on the arc that does not contain , and let and denote the incenters of triangles and , respectively.
Prove that as varies, the circumcircle of triangle passes through a fixed point.
Solution 1
We claim that (midpoint of arc ) is the fixed point. We would like to show that , , , are cyclic.
We extend to intersect again at R. We extend to intersect again at S.
We invert around a circle centered at with radius (for convenience). (I will denote X' as the reflection of X for all the points) The problem then becomes: Prove , , and are collinear.
Now we look at triangle . We apply Menelaus (the version where all three points lie outside the triangle). It suffices to show that
By inversion, we know for any point and for any points and .
Plugging this into our Menelaus equation we obtain that it suffices to show We cancel out the like terms and rewrite. It suffices to show We know that is the diameter of because is isosceles and is the angle bisector. We also know so and are symmetric with respect to so .
Thus, it suffices to show . This is obvious because . Therefore we are done.
Solution 2
We will use complex numbers, with the circumcircle of as the unit circle. Let such that We claim that the circumcircle of passes through This is true iff is real. This is true iff We can compute so we are done.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2016 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |