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. |
+ | 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: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \angle{EDF} &= \angle{EDA} + \angle{ADF} \ | ||
+ | &= \angle{XEA} + \angle{AFX} \ | ||
+ | &= (180^\circ - \angle{AEX}) + (180^\circ - \angle{YFA}) \ | ||
+ | &= 2\angle{FAB} + 2\angle{CAE}\ | ||
+ | &= 2(\angle{FAE} - 90^\circ)\ | ||
+ | &= 2(90^\circ - \angle{EDF}), | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | 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 | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \angle{DYF} &= \angle{FAB} + \angle{BAD} \ | ||
+ | &= \angle{FDY} \ | ||
+ | &= \angle{YFD}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Then, <math>\angle{FED} = 60^\circ</math> as well, and we are done. | ||
==See also== | ==See also== |
Revision as of 15:59, 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 will prove that triangle
is equilateral. To see this, note that
Then,
as well, and we are done.
See also
2006 Canadian MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 | Followed by Last question |