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. $\square$

We prove by contradiction that it cannot happen. Suppose $EF$ is tangent to the A-excircle. Let $\omega$ be the A-excircle. Then, $\omega$ is also the excircle of triangle $AEF$. Because $\triangle{AEF} \sim \triangle{ABC}$, this tells us that $\triangle{ABC}$ is congruent to $\triangle{AEF}$. However, $AE=AB\cos{A}$ which means that $\angle{A}=0$, meaning this is a contradiction, so the proof is complete.

Here's the solution we submitted. It uses the existence of the orthocenter.

If a point is equidistant from two lines, then it must lie on the angle bisector of the angle formed by those two lines. $O$ then has to lie on the angle bisector of $A$, as well as the perpendicular bisector of $BE$, but by angle bisector, these two lines intersect inside the triangle, thus a contradiction.

My solution utilized contradiction with equal tangents and led to $A$, $E$, and $F$ being collinear, which is obviously impossible.

Let the $A$-excircle touch $\overline{AC}$, $\overline{AB}$, $\overline{EF}$ at $B'$, $S$, $C'$. [asy][asy] size(6cm); defaultpen(fontsize(9pt)); pen pri=royalblue; pen sec=Cyan; pen tri=deepgreen; pen fil=invisible; pair A, B, C, EE, IA, Bp, Cp, SS, F; A=dir(140); B=dir(230); C=dir(310); EE=foot(B, A, C); IA=2*dir(270)-incenter(A, B, C); Bp=foot(IA, A, C); Cp=foot(IA, A, B); SS=intersectionpoints(circle(IA, length(Bp-IA)), circle((EE+IA)/2, length(EE-IA)/2))[1]; F=extension(A, B, EE, SS); draw(extension(A, B, IA-(0, length(IA-Bp)), IA-(1, length(IA-Bp))) -- A -- extension(A, C, IA-(0, length(IA-Bp)), IA-(1, length(IA-Bp))), pri); draw(B -- C, pri); filldraw(circle(IA, length(Bp-IA)), fil, pri); fill(A--B--C--cycle,fil); draw(B -- EE, sec); draw(C -- F, sec); draw(EE -- F, tri); draw(Bp -- IA -- Cp, pri); draw(SS -- IA, tri); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, NE); dot("$E$", EE, NE); dot("$I_A$", IA, SE); dot("$B'$", Bp, NE); dot("$C'$", Cp, W); dot("$S$", SS, dir(120)); dot("$F$", F, W); [/asy][/asy] If $\rho$ denotes the semiperimeter, then \begin{align*} BC\cos A&=EF=ES+FS=EB'+FC'\\ &=(\rho-AE)+(\rho-AF)\\ &=AB+BC+CA-(AB+CA)\cos A, \end{align*}from which $\cos A=1$, absurd.

We reflect $BC$ over the $A$ angle bisector to $B'C'$. Evidently $B'C' || EF$ by simple angle chasing. However, $B'C'$ is tangent to the excircle also by symmetry. We have that $\triangle AEF \sim \triangle AB'C'$. But clearly $AB' = AB >AE$ by right triangle $\triangle AEB$, so thus $EF$ is closer to $A$ than $B'C'$ and can't intersect the excircle. Can someone verify this?

So coordinate bash was not official solution...

Again I bash where it isn't necessary... I guess I'm just like that. Basically, I assumed EF was tangent and computed EF twice, once using equal tangents and once using Ptolemy on $BEFC$. Then we finally get $(a+b+c)(a-b+c)(a+b-c) = 0$, impossible.

When you solve 5 and 6 but not 4.

kvedula2004 wrote: So coordinate bash was not official solution... i happened to coordbash too. set $A=(a, 0), B=(b, 0), C=(0, 1)$ with $a>b>0$ get very long polynomial in terms of $a, b$ with positive integer coefficients $=0$ or $a=b$ both are clearly absurd

Thank god the geo problem was only #4.

If one proves that segment EF cannot be tangent to the A-excircle how many points would one get?

SimonSun wrote: If one proves that segment EF cannot be tangent to the A-excircle how many points would one get? 0 becayse that's like obvious

segment? thats like 0

wait wut It's impossible for a point on line EF but not segment EF to touch the A-excircle for obvious reasons I'd say a 6

Uhh.... doesn't the $A$-excircle have to be tangent to segment $EF$? (because the excircle is completely contained within $AB, AC$). Am I missing something? Edit: Honestly it seems so minor that you might get a 7...

alifenix- wrote: Uhh.... doesn't the $A$-excircle have to be tangent to segment $EF$? (because the excircle is completely contained within $AB, AC$). Am I missing something? Yeah because the extensions go past the extensions of the sides.

did anyone else angle chase to get $\angle ACB=90^{\circ}$ which is impossible? praying for no fake solve

JustKeepRunning wrote: How does this imply that $I,B,F,J,C,E$ all lie on one circle? It doesn't My post was just answering why $\angle BJC = 90^{\circ} - \tfrac12\angle A$. For the main proof: assume for contradiction $A$-excircle is tangent to all four lines. Then, point $J$ is on the internal bisectors of $\angle FBC$ and $\angle FEC$. The intersection of these bisectors must be the arc midpoint of $\widehat{FC}$ of the circumcircle of $FBEC$. In particular this implies $BFJCE$ cyclic. The conclusion follows from there.

@CantonMathGuy You most definitely made your figure look like a ice cream cone on purpose .

$AB\cdot AC > AE\cdot AC$. Inversion ($A; \sqrt{AB\cdot AC}$) maps excircle to mixtilinear incircle which is tangent to ($ABC$). So inversion ($A; \sqrt{AE\cdot AC}$) maps excircle to the circle sitting inside ($ABC$) and it also maps $EF$ to ($ABC$). $\blacksquare$

Another simple proof. Since $E$ is on segment $AC$, it is clear that tangency point cannot be outside segment $EF$. But segment $EF$ also can't be tangent to excircle because it's a chord of nine-point circle, which is externally tangent to excircle.

Here's my video explanation of the problem: https://www.youtube.com/watch?v=9y_77B_bfyY

No, $EF$ can't be tangent to the $A$-excircle. Assume that $EF$ is initially tangent. We will show by contradiction that this is never the case. Denote $I_A$ as the $A$-execenter. First note that quadrilateral $BECF$ is cyclic, and the main crux of the problem is proving that $I_A$ lies on the circle that contains this quadrilateral. Claim: $I_A$ lies on the circle that contains quadrilateral $BECF$ Proof: We can start off by noticing that $\triangle AEF$ shares the same excenter as $\triangle ABC$, so $I_AE$ will bisect the exterior angle of $\angle AEF$ and $I_AB$ will bisect the exterior angle of $\angle ABC$. Now $\triangle ABC \sim \triangle AEF$. This is trivial to show, as $\angle BAC = \angle EAF$ and $\angle AFE = \angle BFE = \angle BCE = \angle BCA$ because quadrilateral $BECF$ is cyclic. This means that $\angle FBC = FEC$, and their angle bisectors are also obviously equal. From this we have that quadrilateral $BECI_A$ is cyclic since $\angle I_ABC = \angle I_AEC$. Hence we have shown that $I_A$ lies on the circle that contains quadrilateral $BECF$. Now since $BC$ is the diameter of this circle, $\angle BI_AC = 90$. However in $\triangle BI_AC$, $\angle I_ABC = 90 - \frac{1}{2}\angle ABC$ and $\angle I_ACB = 90 - \frac{1}{2}\angle ACB$, so we have that $\angle BI_AC = 90 - \frac{1}{2}\angle BAC$. The only way $\angle BI_AC = 90$ is if $\angle BAC = 0$, which is impossible. We have reached a contradiction. $\blacksquare$

Oops someone else thought to overkill this problem before me By similar triangles, $AB\cdot AF=AE\cdot AC$. Under an inversion centered at $A$ with power $\sqrt{AB\cdot AC}$, we know that the excircle is swapped with the mixtilinear incircle, so under an inversion centered at $A$ with power $\sqrt{AB\cdot AF}=\sqrt{AE\cdot AC}<\sqrt{AB\cdot AC}$, the excircle's image will be strictly contained within $(ABC)$, while $EF$'s image is $(ABC)$.

Lemma: Triangle $AEF$ is similar to triangle $ABC$. Proof: It suffices to show that $\angle{AFE}=\angle{ACB}$, which is true if $ECFB$ is cyclic. This is true because $\angle{CFB}=\angle{BEC}=90$. $\blacksquare$ Answer is $\boxed{\text{no}}$. Suppose $EF$ is tangent to the $A$-excircle. Then the $A$-excircle of $\triangle ABC$ is the same as the $A$-excircle of $\triangle AEF$. Note that the $A$-exradius of $\triangle ABC$ is equal to $\frac{k}{s-a}$, where $a$ is the length of $BC$, $k$ is the area of $ABC$ and $s$ is the semiperimeter of $ABC$. Let the ratio of sides between triangles $ABC$ and $AEF$ be $x$. So the $A$-exradius of $\triangle AEF$ is $\frac{kx^2}{sx-ax}=\frac{kx}{s-a}$. Thus, $\frac{k}{s-a}=\frac{kx}{s-a}\implies k=kx\implies x=1$. So $\triangle ABC$ is congruent to $\triangle AEF$, which implies $AB=AE$. However, $1=\frac{AE}{AB}=\cos\angle{A}\implies \angle{A}=0$, a contradiction.

The answer is no. Suppose otherwise, so $\triangle ABC$ and $\triangle AEF$ share an $A$-excircle. Further, $BECF$ is cyclic as $\angle BEC=\angle BFC=90^\circ$, so $\angle BFE=\angle BCE \implies \angle AFE=\angle ACB$. As such, $\triangle ABC \sim \triangle AEF$, and since their $A$-exradii are equal they must be congruent. Hence $\overline{EF}$ must be the reflection of $\overline{BC}$ over the $\angle A$-bisector, implying $\triangle ABE$ is isosceles with $AB=AE$. But then $\angle ABE=\angle AEB=90^\circ \implies \angle A=0$, which is absurd. $\blacksquare$

The answer is no. Suppose otherwise; let the excircle be tangent to $\overline{AB}$ and $\overline{AC}$ at two points $X$ and $Y$. Then $$a \cos A = EF=FX+EY=(s-c)+a\cos B + (s-b) + a \cos C = a + a \cos B + a \cos C.$$Then \begin{align*} 1+\cos B + \cos C - \cos A &= 0 \\ (a+b+c)(a-b+c)(a+b-c) &=0, \end{align*}which is impossible by the triangle inequality.

The answer is no. Let $I_a$ denote the $A$-excenter, $R_a$ denote the $A$-exradius, and $l$ be the line passing through $A$ parallel to $BC$. It's well-known that $AA$ and $l$ are isogonal in $\angle BAC$. Moreover, Reim's implies $AA \parallel EF$. Now, since $I_a$ lies on the internal bisector of $\angle BAC$, we have $$\text{dist}(A, EF) + \text{dist}(I_a, EF) = \text{dist}(I_a, AA) = \text{dist}(I_a, l)$$$$= \text{dist}(A, BC) + \text{dist}(I_a, BC) = \text{dist}(A, BC) + R_a.$$It's clear that $BECF$ is cyclic, so $ABC \sim AEF$. This yields $$\frac{\text{dist}(A, EF)}{\text{dist}(A, BC)} = \frac{AE}{AB} = \cos A < 1$$where the last inequality follows from $0^{\circ} < A < 90^{\circ}$. Thus, $$\text{dist}(I_a, EF) - R_a = \text{dist}(A, BC) - \text{dist}(A, EF) > 0$$which finishes. $\blacksquare$ Remarks: I solved this problem on the Monday night before JMO happened this year. I had also returned from Greece the night before. Nevertheless, my solution is incredibly shortsighted, inefficient, and overcomplicated. The best solution is to realize if $EF$ can be tangent to the $A$-excircle, then $AEF$ must have the same $A$-excircle as $ABC$. (Then, you can use $|\cos A| < 1$ to argue that this is impossible.)

Let $T_B$ and $T_C$ be the points at which the $A$-excircle is tangent to the extension of $\overline{AB}$ and the extension of $\overline{AC}$, respectively. Now, assume for contradiction $\overleftrightarrow{EF}$ was tangent to the $A$-excircle at $T$. Note that we have the following equalities: \[ FT = FT_B = BT_B - BF = \frac{a + b - c}{2} + a\cos B,\]\[ ET = ET_C = EC + CT_C = a\cos C + \frac{a + c - b}{2}.\]Also, note that $ECFB$ is cyclic as $\angle BEC = \angle BFC = 90^\circ$. Using Ptolemy's in this quadrilateral gives \[ EF\cdot a = a\sin C\cdot a\sin B - a\cos B\cdot a\cos C = -a^2\cos (B+ C),\]so $EF = -a\cos (B+C)$. Hence, \[ EF = FT + ET\implies -a\cos (B+C) = a + a\cos B + a\cos C\implies \cos (B+C) + \cos B + \cos C + 1 =0.\]Now, let $\cos B = x$ and $\cos C = y$. Then, from the above equation we have \[ xy - \sqrt{(1-x^2)(1-y)^2} + x + y + 1 = 0,\]\[ (x+1)(y+1) = \sqrt{(1-x)(1-y)(1+x)(1+y},\]\[ \sqrt{(x+1)(y+1)} =\sqrt{(1-x)(1-y)}\implies xy + x + y + 1 = xy - x - y + 1\implies x = -y.\]Hence, $\cos B = -\cos C\implies B = 180^\circ - C$. However, this means $A = 0$, a contradiction.

Another question trivialised by Cartesian Coordinates. Set $F=(0,0),A=(a,0),B=(b,0),C=(0,1)$. Bashing will eventually work. Full proof here: https://infinityintegral.substack.com/p/usajmo-2019-contest-review

nop. lets show it cannot. let the excircle intersect line AC at X. Then if EF is tangent to the A-excircle, then $EF \ge EX$ (since tangents are equal length, and EF can be tangent anywhere from E to F.) $EF=BC cosA$ from orthic triangle properties, and $AX=$semiperimeter of $ABC$ by excircle properties, so $EX=AX-AE=AC+AB+BC-ABcosA$ Setting $EF \ge EX$ we get $BC cos A \ge \frac{AB+AC+BC}{2} - ABcosA$ and factoring gives $(2cosA-1)(AB+BC) \ge AC$, and by triangle inequality we have $AB+BC \ge AC$ so we need $(2cosA-1)=1$, so $cosA=1$ which gives $A=0$ degrees which is impossible.

If we assume that condition is true, there is two different inversions have to be same. (Here how it goes like). (1) Inverting through the circle centered at $A$ and radius $AB*AF$. Then EF transforms into $(ABC)$. And since lines $AB$ and $AC $ remains same after inversion, also image of excircles under inversion must be tangent to these lines. If we assume &EF& was tangent to A-excircle then it's image must be tangent to $(ABC)$. Which means that image of A-excircle must be A-mixtilinear circle of $ABC$. (2) Inverting through the circle centered at $A$ and radius $AB*AC$. Then A-excircle and A-mixtilinear circles have to be inversion pair. Combining (1) and (2) we get that these two inversions are same, which implies these two have same radius which is clearly impossible since $AC>AF$