2019 USAJMO Problems/Problem 4

$(*)$ Let $ABC$ be a triangle with $\angle ABC$ obtuse. The [i] $A$ -excircle[/i] is a circle in the exterior of $\triangle ABC$ that is tangent to side $\overline{BC}$ of the triangle and tangent to the extensions of the other two sides. Let $E$ , $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$ , respectively. Can line $EF$ be tangent to the $A$ -excircle?

Solution

Instead of trying to find a synthetic way to describe $EF$ being tangent to the $A$ -excircle (very hard), we instead consider the foot of the perpendicular from the $A$ -excircle to $EF$ , hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe $EF$ , something more closely related to the $A$ -excircle; as we are considering perpendicularity, if we could generate a line parallel to $EF$ , that would be good.

So we recall that it is well known that triangle $AEF$ is similar to $ABC$ . This motivates reflecting $BC$ over the angle bisector at $A$ to obtain $B'C'$ , which is parallel to $EF$ for obvious reasons.

Furthermore, as reflection preserves intersection, $B'C'$ is tangent to the reflection of the $A$ -excircle over the $A$ -angle bisector. But it is well-known that the $A$ -excenter lies on the $A$ -angle bisector, so the $A$ -excircle must be preserved under reflection over the $A$ -excircle. Thus $B'C'$ is tangent to the $A$ -excircle.Yet for all lines parallel to $EF$ , there are only two lines tangent to the $A$ -excircle, and only one possibility for $EF$ , so $EF = B'C'$ .

Thus as $ABB'$ is isoceles, $[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,$ 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(" $A$ ", A, dir(110)); dot(" $B$ ", B, dir(190)); dot(" $C$ ", C, dir(50)); dot(" $E$ ", E, dir(50)); dot(" $F$ ", F, dir(190)); dot(" $E'$ ", Ep, dir(190)); dot(" $F'$ ", Fp, dir(50)); [/asy]

Suppose otherwise. Consider the reflection over the bisector of $\angle BAC$ . This swaps rays $AB$ and $AC$ ; suppose $E$ and $F$ are sent to $E'$ and $F'$ . Note that the $A$ -excircle is fixed, so line $E'F'$ must also be tangent to the $A$ -excircle.

Since $BEFC$ is cyclic, we obtain $\measuredangle ECB = \measuredangle EFB = \measuredangle EF'E'$ , so $\overline{E'F'} \parallel \overline{BC}$ . However, as $\overline{EF}$ is a chord in the circle with diameter $\overline{BC}$ , $EF \le BC$ .

If $EF < BC$ then $E'F' < BC$ too, so then $\overline{E'F'}$ lies inside $\triangle ABC$ and cannot be tangent to the excircle.

The remaining case is when $EF = BC$ . In this case, $\overline{EF}$ is also a diameter, so $BECF$ is a rectangle. In particular $\overline{BE} \parallel \overline{CF}$ . However, by the existence of the orthocenter, the lines $BE$ and $CF$ must intersect, contradiction.

2019 USAJMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
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2019 USAJMO Problems/Problem 4

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Solution 2

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