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*} | ||
Line 31: | Line 34: | ||
\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 16:02, 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 |