Difference between revisions of "1985 IMO Problems/Problem 1"
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=== Solution 1 === | === Solution 1 === | ||
− | Let <math> | + | Let <math>O</math> be the center of the circle mentioned in the problem. Let <math>T</math> be the second intersection of the circumcircle of <math>CDO </math> with <math>AB </math>. By measures of arcs, <math> \angle DTA = \angle DCO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2} </math>. It follows that <math>AT = AD </math>. Likewise, <math>TB = BC</math>, so <math>AD + BC = AB </math>, as desired. |
=== Solution 2 === | === Solution 2 === |
Revision as of 15:18, 3 April 2016
Problem
A circle has center on the side of the cyclic quadrilateral
. The other three sides are tangent to the circle. Prove that
.
Solutions
Solution 1
Let be the center of the circle mentioned in the problem. Let
be the second intersection of the circumcircle of
with
. By measures of arcs,
. It follows that
. Likewise,
, so
, as desired.
Solution 2
Let be the center of the circle mentioned in the problem, and let
be the point on
such that
. Then
, so
is a cyclic quadrilateral and
is in fact the
of the previous solution. The conclusion follows.
Solution 3
Let the circle have center and radius
, and let its points of tangency with
be
, respectively. Since
is clearly a cyclic quadrilateral, the angle
is equal to half the angle
. Then
Likewise, . It follows that
,
Q.E.D.
Solution 4
We use the notation of the previous solution. Let be the point on the ray
such that
. We note that
;
; and
; hence the triangles
are congruent; hence
and
. Similarly,
. Therefore
, Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1985 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |