2006 Canadian MO Problems/Problem 5


The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $AB$ and arc $AC$ are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines $AB$ and $AC$. More precisely, the point $D$ on arc $BC$ is the midpoint of the segment joining the points $D^\prime$ and $D^\prime^\prime$ (Error compiling LaTeX. Unknown error_msg) where the tangent at $D$ intersects the extended lines $AB$ and $AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$. Prove that triangle $DEF$ is equilateral.


Let the intersection of the tangents at $D$ and $E$, $E$ and $F$, $F$ and $D$ be labeled $Z, X,Y$, respectively.

It is a well-known fact that in a right triangle $PQR$ with $M$ the midpoint of hypotenuse $PR$, triangles $MQR$ and $PQM$ are isosceles.

Now we do some angle-chasing: \begin{align*} \angle{EDF} &= \angle{EDA} + \angle{ADF} \\                      &= \angle{XEA} + \angle{AFX} \\                      &= (180^\circ - \angle{AEZ}) + (180^\circ - \angle{YFA}) \\                      &= 2\angle{FAB} + 2\angle{CAE}\\                      &= 2(\angle{FAE} - 90^\circ)\\                      &= 2(90^\circ - \angle{EDF}), \end{align*} whence we conclude that $\angle{EDF} = 60^\circ.$

Next, we prove that triangle $DYF$ is equilateral. To see this, note that \begin{align*} \angle{DYF} &= \angle{FAB} + \angle{BAD} \\                      &= \angle{FDY} \\                      &= \angle{YFD}. \end{align*} Hence $\angle{FED} = 60^\circ$ as well, so triangle $DEF$ is equilateral as desired.


See also

2006 Canadian MO (Problems)
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Problem 4
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