Difference between revisions of "2019 USAJMO Problems/Problem 4"
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pair A, B, C, E, F, Ep, Fp, I, J, K, L; | pair A, B, C, E, F, Ep, Fp, I, J, K, L; |
Revision as of 18:09, 19 April 2019
Let be a triangle with obtuse. The [i]-excircle[/i] is a circle in the exterior of that is tangent to side of the triangle and tangent to the extensions of the other two sides. Let , be the feet of the altitudes from and to lines and , respectively. Can line be tangent to the -excircle?
Solution
Instead of trying to find a synthetic way to describe being tangent to the -excircle (very hard), we instead consider the foot of the perpendicular from the -excircle to , hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe , something more closely related to the -excircle; as we are considering perpendicularity, if we could generate a line parallel to , that would be good.
So we recall that it is well known that triangle is similar to . This motivates reflecting over the angle bisector at to obtain , which is parallel to for obvious reasons.
Furthermore, as reflection preserves intersection, is tangent to the reflection of the -excircle over the -angle bisector. But it is well-known that the -excenter lies on the -angle bisector, so the -excircle must be preserved under reflection over the -excircle. Thus is tangent to the -excircle.Yet for all lines parallel to , there are only two lines tangent to the -excircle, and only one possibility for , so .
Thus as is isoceles, contradiction. -alifenix-
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution 2
The answer is no.
[asy] pair A, B, C, E, F, Ep, Fp, I, J, K, L; //A = dir(120); //B = 2*foot(origin, A, A+dir(250)) - A; //C = 2*foot(origin, A, A+dir(290)) - A; A = dir(155); B = dir(230); C = dir(310); E = foot(B, A, C); F = foot(C, A, B); I = incenter(A, B, C); J = 2*circumcenter(B, I, C) - I; K = foot(J, A, B); L = foot(J, A, C); Ep = reflect(A, I) * E; Fp = reflect(A, I) * F;
draw(B--K^^C--L, rgb(1, 0.7, 0.2)); //draw(unitcircle, dotted); draw(A--B^^A--C, rgb(0.7, 0.3, 0)); draw(B--C, heavycyan); draw(E--F, heavycyan); draw(Ep--Fp, heavycyan + dashed); filldraw(circle(J, abs(J-foot(J, B, C))), opacity(0.5) + pink, red); draw(B--E^^C--F, heavymagenta); draw(E--Ep^^F--Fp, dotted);
dot("", A, dir(110)); dot("", B, dir(190)); dot("", C, dir(50)); dot("", E, dir(50)); dot("", F, dir(190)); dot("", Ep, dir(190)); dot("", Fp, dir(50)); [/asy]
Suppose otherwise. Consider the reflection over the bisector of . This swaps rays and ; suppose and are sent to and . Note that the -excircle is fixed, so line must also be tangent to the -excircle.
Since is cyclic, we obtain , so . However, as is a chord in the circle with diameter , .
If then too, so then lies inside and cannot be tangent to the excircle.
The remaining case is when . In this case, is also a diameter, so is a rectangle. In particular . However, by the existence of the orthocenter, the lines and must intersect, contradiction.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |