Difference between revisions of "2003 USAMO Problems/Problem 4"
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Combining the above, we conclude that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | Combining the above, we conclude that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>. | ||
+ | |||
+ | |||
+ | === Solution 3 === | ||
+ | We will construct a series of equivalent statements. First: | ||
+ | <cmath>\begin{align} | ||
+ | MB\cdot MD = MC^2 &\equiv \frac{MB}{MC} = \frac{MC}{MD} \ | ||
+ | &\equiv \triangle MCD\sim \triangle MBC \ | ||
+ | &\equiv \angle CBA = \angle FCA | ||
+ | \end{align}</cmath> | ||
+ | where the second equivalence follows from SAS similarity (because <math>\angle CMD = \angle BMC</math>). | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 22:06, 5 April 2016
Problem
Let be a triangle. A circle passing through
and
intersects segments
and
at
and
, respectively. Lines
and
intersect at
, while lines
and
intersect at
. Prove that
if and only if
.
Solutions
Solution 1
Extend segment through
to
such that
.

Then if and only if quadrilateral
is a parallelogram, or,
. Hence
if and only if
, that is,
.
Because quadrilateral is cyclic,
. It follows that
if and only if
that is, quadrilateral
is cyclic, which is equivalent to
Because
,
if and only if triangles
and
are similar, that is
or
.
Solution 2
We first assume that . Because
and
, triangles
and
are similar. Consequently,
.

Because quadrilateral is cyclic,
. Hence
implying that
, so
. Because quadrilateral
is cyclic,
. Hence
Because
and
, triangles
and
are similar. Consequently,
, or
. Therefore
implies
.
Now we assume that . Applying Ceva's Theorem to triangle
and cevians
gives
implying that
, so
.
Consequently, . Because quadrilateral
is cyclic,
. Hence
Because
and
, triangles
and
are similar. Consequently,
, or
.
Combining the above, we conclude that if and only if
.
Solution 3
We will construct a series of equivalent statements. First:
where the second equivalence follows from SAS similarity (because
).
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.