Difference between revisions of "2013 USAMO Problems/Problem 1"
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using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>. | using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>. | ||
+ | ==Solution 4== | ||
+ | We can use some construction arguments to solve the problem. | ||
+ | |||
+ | Let <math>\angle BAC=\alpha,</math> <math>\angle ABC=\beta,</math> <math>\angle ACB=\gamma,</math> and let <math>\angle APB=\theta.</math> We construct lines through the points <math>Q,</math> and <math>R</math> that intersect with <math>\triangle ABC</math> at the points <math>Q</math> and <math>R,</math> respectively, and that intersect each other at <math>T.</math> We will construct these lines such that <math>\angle CQV=\angle ARV=\theta.</math> | ||
+ | |||
+ | |||
+ | Now we let the intersections of <math>AP</math> with <math>QU</math> and <math>RV</math> be <math>Y'</math> and <math>Z',</math> respectively. This construction is as follows. | ||
+ | <asy> | ||
+ | import olympiad; | ||
+ | import cse5; | ||
+ | size(11cm); | ||
+ | real lsf=0.8000; | ||
+ | real lisf=2011.0; | ||
+ | |||
+ | defaultpen(fontsize(10pt)); | ||
+ | |||
+ | pair A = (-1.0, 3.0); | ||
+ | pair B = (-3.0, -3.0); | ||
+ | pair C = (4.0, -3.0); | ||
+ | pair P = (-0.6698198198198195, -3.0); | ||
+ | pair Q = (1.1406465288818244, 0.43122416534181074); | ||
+ | pair R = (-1.6269590345062048, 1.119122896481385); | ||
+ | path w_A = circumcircle(A,Q,R); | ||
+ | path w_B = circumcircle(B,P,R); | ||
+ | path w_C = circumcircle(P,Q,C); | ||
+ | pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5)); | ||
+ | pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5)); | ||
+ | pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5)); | ||
+ | pair X = (2)*(foot(O_A,A,P))-A; | ||
+ | pair Y = (2)*(foot(O_B,A,P))-P; | ||
+ | pair Z = (2)*(foot(O_C,A,P))-P; | ||
+ | pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P; | ||
+ | pair D = (2)*(foot(O_B,X,M))-M; | ||
+ | pair E = (2)*(foot(O_C,X,M))-M; | ||
+ | |||
+ | draw(A--B--C--A, rgb(0.6,0.6,0.0)); | ||
+ | draw(A--P, rgb(0.0,0.2,0.4)); | ||
+ | draw(P--M, rgb(0.4,0.2,0.0)); | ||
+ | draw(R--M, rgb(0.4,0.2,0.0)); | ||
+ | draw(Q--M, rgb(0.4,0.2,0.0)); | ||
+ | draw(B--M, rgb(0.0,0.2,0.4)); | ||
+ | draw(C--M, rgb(0.0,0.2,0.4)); | ||
+ | draw(R--Y); | ||
+ | draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8)); | ||
+ | draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8)); | ||
+ | draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8)); | ||
+ | draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); | ||
+ | |||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(P); | ||
+ | dot(Q); | ||
+ | dot(R); | ||
+ | dot(X); | ||
+ | dot(Y); | ||
+ | dot(Z); | ||
+ | dot(M); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | label("$A$", A, lsf * dir(110)); | ||
+ | label("$B$", B, lsf * unit(B-M)); | ||
+ | label("$C$", C, lsf * unit(C-M)); | ||
+ | label("$P$", P, lsf * unit(P-M) * 1.8); | ||
+ | label("$Q$", Q, lsf * dir(90) * 1.6); | ||
+ | label("$R$", R, lsf * unit(R-M) * 2); | ||
+ | label("$X$", X, lsf * dir(-60) * 2); | ||
+ | label("$Y$", Y, lsf * dir(45)); | ||
+ | label("$Z$", Z, lsf * dir(5)); | ||
+ | label("$M$", M, lsf * dir(M-P)*2); | ||
+ | label("$D$", D, lsf * dir(150)); | ||
+ | label("$E$", E, lsf * dir(5)); | ||
+ | </asy> | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:02, 4 August 2020
Problem
In triangle , points
lie on sides
respectively. Let
,
,
denote the circumcircles of triangles
,
,
, respectively. Given the fact that segment
intersects
,
,
again at
respectively, prove that
Solution 1
In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that concur at a point
. Let
meet
again at
and
, respectively. Then by Power of a Point, we have
Thusly
But we claim that
. Indeed,
and
Therefore,
. Analogously we find that
and we are done.
courtesy v_enhance
Solution 2
Diagram Refer to the Diagram link.
By Miquel's Theorem, there exists a point at which intersect. We denote this point by
Now, we angle chase:
In addition, we have
Now, by the Ratio Lemma, we have
(by the Law of Sines in
)
(by the Law of Sines in
)
by the Ratio Lemma.
The proof is complete.
Solution 3
Use directed angles modulo .
Lemma.
Proof.
Now, it follows that (now not using directed angles)
using the facts that
and
,
and
are similar triangles, and that
equals twice the circumradius of the circumcircle of
.
Solution 4
We can use some construction arguments to solve the problem.
Let
and let
We construct lines through the points
and
that intersect with
at the points
and
respectively, and that intersect each other at
We will construct these lines such that
Now we let the intersections of with
and
be
and
respectively. This construction is as follows.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.