Difference between revisions of "2019 USAJMO Problems/Problem 4"

Problem

$(*)$ Let $ABC$ be a triangle with $\angle ABC$ obtuse. The $A$-excircle 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-

Solution 2

The answer is no.

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.

Solution 3

The answer is $\boxed{\text{no}}$.

Suppose for the sake of contradiction that it is possible for $EF$ to be tangent to the $A$-excircle. Call the tangency point $T$, and let $S_1, S_2$ denote the contact points of $AB, AC$ with the $A$-excircle, respectively. Let $s$ denote the semiperimeter of $ABC$. By equal tangents, we have $$ET = ES_2, FT = FS_1 \implies EF = ES_2+FS_2$$It is also well known that $AS_1 = AS_2 = \frac{s}{2}$, so $$EF = ES_2+FS_2 = (AS_2-AE)+(AS_1-AF) = s-AE-AF \implies s=AE+AF+EF$$It is well known (by an easy angle chase) that $\triangle AEF \sim \triangle ABC$, so we must have the ratio of similitude is $2$. In particular, $$AB=2 \cdot AE, AC=2 \cdot AF$$This results in $$\angle ABE = 30^{\circ}, \angle CBF = 30^{\circ} \implies \angle EBC = 120^{\circ}$$which is absurd since $\triangle BEC$ is a right triangle. We reached a contradiction, so we are done.