Difference between revisions of "2016 USAJMO Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
+ | |||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A = dir(90); | ||
+ | pair B = dir(-10); | ||
+ | pair C = dir(190); | ||
+ | pair P = dir(-70); | ||
+ | pair U = incenter(A,B,P); | ||
+ | pair V = incenter(A,C,P); | ||
+ | pair M = dir(-90); | ||
+ | |||
+ | draw(circle((0,0),1)); | ||
+ | dot("$A$", A, dir(A)); | ||
+ | dot("$B$", B, dir(B)); | ||
+ | dot("$C$", C, dir(C)); | ||
+ | dot("$P$", P, dir(P)); | ||
+ | dot("$I_B$", U, NE); | ||
+ | dot("$I_C$", V, NW); | ||
+ | dot("$M$", M, dir(M)); | ||
+ | draw(A--B--C--A); | ||
+ | draw(circumcircle(P,U,V)); | ||
+ | |||
+ | |||
+ | </asy> | ||
We claim that <math>M</math> (midpoint of arc <math>BC</math>) is the fixed point. | We claim that <math>M</math> (midpoint of arc <math>BC</math>) is the fixed point. | ||
Line 17: | Line 41: | ||
The problem then becomes: Prove <math>I_B'</math>, <math>I_C'</math>, and <math>M'</math> are collinear. | 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 | + | Now we look at triangle <math>\triangle PR'S'</math>. We apply Menelaus (the version where all three points lie outside the triangle). |
It suffices to show that | It suffices to show that | ||
− | <cmath>\dfrac{ | + | <cmath>\dfrac{PI_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> | + | By inversion, we know <math>PX' = \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 | Plugging this into our Menelaus equation we obtain that it suffices to show | ||
Line 31: | Line 55: | ||
Thus, it suffices to show <math>\dfrac{SI_C}{RI_B} = 1</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>. | This is obvious because <math>RI_B = RA = SA = SI_C</math>. | ||
− | Therefore we are done. <math>\ | + | Therefore we are done. <math>\blacksquare</math> |
== Solution 2== | == Solution 2== | ||
− | We will use complex numbers | + | We will use complex numbers as mentioned [http://web.mit.edu/yufeiz/www/wc08/peng_formula.pdf here]. Set the circumcircle of <math>\triangle ABC</math> to be 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 if <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. Now observe that <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 <math>k</math> is real and we are done. <math>\blacksquare</math> |
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A = dir(90); | ||
+ | pair B = dir(-10); | ||
+ | pair C = dir(190); | ||
+ | pair P = dir(-70); | ||
+ | pair U = incenter(A,B,P); | ||
+ | pair V = incenter(A,C,P); | ||
+ | pair M = dir(-90); | ||
+ | pair D = dir(40); | ||
+ | pair E = dir(140); | ||
+ | |||
+ | |||
+ | draw(circle((0,0),1)); | ||
+ | dot("$A$", A, dir(A)); | ||
+ | dot("$B$", B, dir(B)); | ||
+ | dot("$C$", C, dir(C)); | ||
+ | dot("$D$", D, dir(D)); | ||
+ | dot("$E$", E, dir(E)); | ||
+ | dot("$P$", P, dir(P)); | ||
+ | dot("$I_B$", U, NE); | ||
+ | dot("$I_C$", V, NW); | ||
+ | dot("$M$", M, dir(M)); | ||
+ | draw(A--B--C--A--P--B ^^ P--C ^^ E--D--P--E--M--V ^^ D--M--U--V); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>M</math> be the midpoint of arc <math>BC</math>. Let <math>D</math> be the midpoint of arc <math>AB</math>. Let <math>E</math> be the midpoint of arc <math>AC</math>. | ||
+ | Then, <math>P, I_B</math>, and <math>D</math> are collinear and <math>P, I_C</math>, and <math>E</math> are collinear. | ||
+ | |||
+ | We'll prove <math>MPI_B I_C</math> is cyclic. (Intuition: we'll show that <math>M</math> is the Miquel's point of quadrilateral <math>DE I_C I_B</math>. | ||
+ | |||
+ | <math>D</math> is the center of the circle <math>A I_B B</math> (the <math>P-</math> excenter of <math>PAB</math> is also on the same circle). Therefore <math>D I_B = DB</math>. Similarly <math>E I_C = EC</math>. Since <math>AB=AC</math>, <math>DB=EC</math>. Therefore <math>D I_B = E I_C</math>. Obviously <math>ME = MD</math> and <math>\angle MEI_C = \angle MEP = \angle MDP = \angle MDI_B</math>. Thus by SAS, <math>\triangle MEI_C \cong \triangle MDI_B</math>. | ||
+ | |||
+ | Hence <math>\angle I_B M I_C = \angle DME = \angle DPE = \angle I_B P I_C</math>, so <math>MPI_B I_C</math> is cyclic and we are done. | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:15, 28 June 2020
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 as mentioned here. Set the circumcircle of to be the unit circle. Let such that We claim that the circumcircle of passes through This is true if is real. Now observe that so is real and we are done.
Solution 3
Let be the midpoint of arc . Let be the midpoint of arc . Let be the midpoint of arc . Then, , and are collinear and , and are collinear.
We'll prove is cyclic. (Intuition: we'll show that is the Miquel's point of quadrilateral .
is the center of the circle (the excenter of is also on the same circle). Therefore . Similarly . Since , . Therefore . Obviously and . Thus by SAS, .
Hence , so is cyclic and 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 |