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Difference between revisions of "2006 Canadian MO Problems/Problem 5"

(→‎Solution)
(→‎Solution)
 
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==Solution==
 
==Solution==
 
Let the intersection of the tangents at <math>D</math> and <math>E</math>, <math>E</math> and <math>F</math>, <math>F</math> and <math>D</math> be labeled <math>Z, X,Y</math>, respectively.
 
Let the intersection of the tangents at <math>D</math> and <math>E</math>, <math>E</math> and <math>F</math>, <math>F</math> and <math>D</math> be labeled <math>Z, X,Y</math>, respectively.
 +
 
It is a well-known fact that in a right triangle <math>PQR</math> with <math>M</math> the midpoint of hypotenuse <math>PR</math>, triangles <math>MQR</math> and <math>PQM</math> are isosceles.
 
It is a well-known fact that in a right triangle <math>PQR</math> with <math>M</math> the midpoint of hypotenuse <math>PR</math>, triangles <math>MQR</math> and <math>PQM</math> are isosceles.
−
Now, we do some angle-chasing:
+
 
 +
Now we do some angle-chasing:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
 
\angle{EDF} &= \angle{EDA} + \angle{ADF} \\
 
\angle{EDF} &= \angle{EDA} + \angle{ADF} \\
 
                     &= \angle{XEA} + \angle{AFX} \\
 
                     &= \angle{XEA} + \angle{AFX} \\
−
                     &= (180^\circ - \angle{AEX}) + (180^\circ - \angle{YFA}) \\
+
                     &= (180^\circ - \angle{AEZ}) + (180^\circ - \angle{YFA}) \\
 
                     &= 2\angle{FAB} + 2\angle{CAE}\\
 
                     &= 2\angle{FAB} + 2\angle{CAE}\\
 
                     &= 2(\angle{FAE} - 90^\circ)\\
 
                     &= 2(\angle{FAE} - 90^\circ)\\
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</cmath>
 
</cmath>
 
whence we conclude that <math>\angle{EDF} = 60^\circ.</math>
 
whence we conclude that <math>\angle{EDF} = 60^\circ.</math>
−
Next, we will prove that triangle<math>DYF</math> is equilateral. To see this, note that
+
 
 +
Next, we prove that triangle <math>DYF</math> is equilateral. To see this, note that
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
Then, <math>\angle{FED} = 60^\circ</math> as well, and we are done.
+
Hence <math>\angle{FED} = 60^\circ</math> as well, so triangle <math>DEF</math> is equilateral as desired.
 +
 
 +
<math>\blacksquare</math>
  
 
==See also==
 
==See also==

Latest revision as of 16:02, 3 June 2011

Problem

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.

Solution

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.

$\blacksquare$

See also

2006 Canadian MO (Problems)
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Problem 4 		1 • 2 • 3 • 4 • 5 Followed by
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