# 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

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. $\blacksquare$ ~ Mathscienceclass

## Solution 4

[asy]

/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */


import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.92, xmax = 28.12, ymin = -14.66, ymax = 5.16; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902);

/* draw figures */


draw((3.52,1.38)--(5.8,-4.7), linewidth(1) + rvwvcq); draw((5.8,-4.7)--(10.3,-4.9), linewidth(1) + wvvxds); draw((10.3,-4.9)--(3.52,1.38), linewidth(1) + rvwvcq); draw(circle((9.325581455949688,-7.26800412402583), 2.399418339060914), linewidth(1) + sexdts); draw((5.8,-4.7)--(7.0789360558927,-8.1104961490472), linewidth(1) + rvwvcq); draw((10.3,-4.9)--(10.956076160142976,-5.507692962492314), linewidth(1) + rvwvcq); draw((8.503617697888226,-3.2360942688404215)--(6.711506849315068,-7.130684931506849), linewidth(1) + wvvxds);

/* dots and labels */


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/* end of picture */


[/asy] We claim that the answer is no. We proceed with contradiction. Suppose that $EF$ is indeed tangent to the a-excenter. Define the point of tangency to be $X$. Let $G$ to be the intersection of the a-excircle with the extension of $AB$ and $H$ to be the intersection of the a-excircle with the extension of $AC$. Define $I_a$ to be the a-excenter. It is a well-known fact that $I_a$ lies on the angle bisector of $\angle BAC$. For convenience, let $AB = c, AC = b, BC = a$, $\angle CAB = A, \angle CBA = B, \angle ACB = C$. Notice that by Power of a Point: $$FX = FG.$$ $$XE = EH.$$ Therefore, adding, we see that: $$FE = FX + XE = FG + EH.$$ It would be rather nice if we could re-write $FG$. Indeed, we can: $$FG = AG - FG = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - b \cos (A)$$ and similarly for $FH$: $$FH = AH - AF = \frac{r_a}{\tan \left( \frac{A}{2} \right)} - c \cos (A)$$ We now seek to re-write $FE$. Using the law of cosines: $$FE^2 = AF^2 + AE^ 2- 2AF \cdot AE \cos (A) = \left(c\cos (A) \right)^2 + \left(b\cos (A) \right)^2 - 2 (bc \cos ^2 (A)) \cos (A) = \cos ^2 (A) (c^2 + b^2 - 2bc \cos (A)) = \cos ^2 (A) \cdot a^2$$ Therefore, $$FE = a \cos (A)$$ Putting this all together, we see that: $$a \cos (A) = \frac{r_a}{\tan \left(\frac{a}{2} \right)} - b \cos (A) + \frac{r_a}{\tan \left(\frac{a}{2} \right)} - c \cos (A) \Longleftrightarrow$$ $$(a+b+c) \cos (A) = \frac{2r_a}{\tan \left(\frac{a}{2} \right)}$$ Now, we seek to write $\tan \left(\frac{a}{2} \right)$ in terms of the side lengths of the triangle. Notice that: $$2\cos ^2 \left(\frac{A}{2} \right)-1 = \cos (A) \Longleftrightarrow$$ $$2\cos ^2 \left(\frac{A}{2} \right) = 1 + \frac{b^2 + c^2-a^2}{2bc} = \frac{b^2 + 2bc + c^2 - a^2}{2bc} = \frac{(b+c)^2-a^2}{2bc} = \frac{(a+b+c)(b+c-a)}{2bc} = \frac{(2s)(2s-2a)}{2bc}$$ where $s$ is the semi-perimeter. We get that: $$\cos \left(\frac{A}{2} \right) = \sqrt{\frac{s(s-a)}{bc}}$$ Using the fact that $\cos^2 \left(\frac{A}{2} \right) + \sin^2 \left(\frac{A}{2} \right) = 1$, we have that: $$\sin^2 \left(\frac{A}{2} \right) = 1-\frac{b^2 + 2bc + c^2 - a^2}{4bc} = \frac{(b-c)^2 - a^2}{4bc} = \frac{(2s-2c)(2s-2b)}{4bc}$$ Therefore, $$\tan \left( \frac{A}{2} \right) = \sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }$$ Returning to our original problem: $$\frac{2r_a}{\sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }} = (a+b+c) \cos (A)$$ It is a well-known fact that $r_a = \sqrt{\frac{s(s-b)(s-c)}{s-a}}$, so: $$\frac{2\sqrt{\frac{s(s-b)(s-c)}{s-a}}}{\sqrt{ \frac{(s-b)(s-c)}{s(s-a)} }} = 2s \cos (A)$$ which implies that: $$\cos (A) =1 \Longleftrightarrow A = 0$$ which is a contradiction. Hence, our original assumption that $EF$ is tangent to the a-excircle is incorrect. $\blacksquare$ ~AopsUser101