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.

2019 USAJMO (Problems • Resources)
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2019 USAJMO Problems/Problem 4

Contents

Problem

Solution

Solution 2

Solution 3

See also