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Greedy Geoff sawed off a corner of a brick shaped block of Christmas cake, exposing a triangular fresh face of moist rich delicious gateau. He placed the tetrahedral fragment on the table, with its fresh face downwards. He mused through a port laden haze that it looked very stable, just like a mountain in fact, with its summit above a point inside its (not necessarily equilateral) triangular footprint $ABC$. He decided to decorate it, and took a UKMT pennant flying from a toothpick, and stuck it at the summit, with the flagpole perfectly vertical. Of course, the port was still at work and he is a bit clumsy, so he jammed the toothpick right through the cake, stabbing it into the tablecloth at a point $X$. Show that the circles $ABX$, $BCX$ and $CAX$ all have the same radius.

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I thought gato meant cat. – Will Jagy Nov 22 '12 at 20:26
Damn it, contest problem: – Will Jagy Nov 22 '12 at 21:50
Could you please make the title a little more descriptive? – Rahul Nov 23 '12 at 3:19
But of course the OP should say (if he knows) "contest problem from 2005". Then Will (and others, like me) won't get all excited about it's appropriateness for this forum. – GEdgar Nov 23 '12 at 16:56

Call summit vertex $S$.

Now, put the piece of cake back where it came from --use the toothpick to hold it in place-- and let's assume that vertex $A$ lies along the vertical edge of the cake. Take the knife and make a vertical cut along $SA$ that slices perpendicularly through $BC$ at, say, $F$. The plane of the cut, being perpendicular to a line in face $ABC$, is necessarily perpendicular to face $ABC$ itself; the cut must have split the toothpick. Therefore, the projection into face $ABC$ of segment $SA$ --together with the projection of $SF$-- forms an altitude of $\triangle ABC$. The same is true of similar cuts through $SB$ and $SC$: the foot of the toothpick lies on all three altitudes, so that $X$ must be the orthocenter of $\triangle ABC$.

From here, proof is fairly straightforward using the plane geometry of $\triangle ABC$: With $X$ the common point on altitudes dropped from $A$, $B$, and $C$, one can show (for instance) that $\angle BXC = 180^\circ-\angle BAC$. Consequently, by the Law of Sines in $\triangle ABC$ and $\triangle XBC$,

$$\text{circumdiameter of } \triangle ABC = \frac{|BC|}{\sin\angle BAC} = \frac{|BC|}{\sin\angle BXC} = \text{circumdiameter of } \triangle XBC$$

Thus, not only are the circumcircles of $\triangle XBC$, $\triangle XCA$, and $\triangle XAB$ congruent to each other, they're congruent to the circumcircle of $\triangle ABC$ itself.

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