Difference between revisions of "2015 USAJMO Problems/Problem 5"
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== Solution 2== | == Solution 2== | ||
− | Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math> | + | All angles are directed. Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math> and <math>CD, CE</math> are isogonal in <math>\triangle CDB</math>. From the law of sines it follows that |
+ | |||
+ | <cmath>\frac{DX}{XB}\cdot \frac{DE}{ED}=\left(\frac{AD}{DB}\right)^2=\left(\frac{DC}{BC}\right)^2.</cmath> | ||
+ | |||
+ | Therefore, the ratio equals <math>\frac{AD\cdot DC}{DB\cdot BC}.</math> | ||
+ | |||
+ | Now let <math>Y</math> be a point of <math>AC</math> such that <math>\angle{ABE}=\angle{CBY}</math>. We apply the above identities for <math>Y</math> to get that <math>\frac{CY}{YC}\cdot \frac{CE}{EA}=\left(\frac{CD}{DA}\right)^2</math>. So <math>\angle{CDY}=\angle{EDA}</math>, the converse follows since all our steps are reversible. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:24, 13 April 2017
Contents
[hide]Problem
Let be a cyclic quadrilateral. Prove that there exists a point
on segment
such that
and
if and only if there exists a point
on segment
such that
and
.
Solution
Note that lines are isogonal in
, so an inversion centered at
with power
composed with a reflection about the angle bisector of
swaps the pairs
and
. Thus,
so that
is a harmonic quadrilateral. By symmetry, if
exists, then
. We have shown the two conditions are equivalent, whence both directions follow
Solution 2
All angles are directed. Note that lines are isogonal in
and
are isogonal in
. From the law of sines it follows that
Therefore, the ratio equals
Now let be a point of
such that
. We apply the above identities for
to get that
. So
, the converse follows since all our steps are reversible.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2015 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |