Difference between revisions of "2006 Canadian MO Problems/Problem 5"
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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 | + | |
+ | 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{ | + | &= (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)\\ | ||
Line 23: | Line 25: | ||
</cmath> | </cmath> | ||
whence we conclude that <math>\angle{EDF} = 60^\circ.</math> | whence we conclude that <math>\angle{EDF} = 60^\circ.</math> | ||
− | Next, we | + | |
+ | 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> | ||
− | + | 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 17:02, 3 June 2011
Problem
The vertices of a right triangle inscribed in a circle divide the circumference into three arcs. The right angle is at , so that the opposite arc is a semicircle while arc and arc 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 and . More precisely, the point on arc is the midpoint of the segment joining the points and $D^\prime^\prime$ (Error compiling LaTeX. Unknown error_msg) where the tangent at intersects the extended lines and . Similarly for on arc and on arc . Prove that triangle is equilateral.
Solution
Let the intersection of the tangents at and , and , and be labeled , respectively.
It is a well-known fact that in a right triangle with the midpoint of hypotenuse , triangles and are isosceles.
Now we do some angle-chasing: whence we conclude that
Next, we prove that triangle is equilateral. To see this, note that Hence as well, so triangle is equilateral as desired.
See also
2006 Canadian MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 | Followed by Last question |