Difference between revisions of "1985 IMO Problems/Problem 1"
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Lemma. Let <math>I</math> be the in-center of <math>ABC</math> and points <math>P</math> and <math>Q</math> be on the lines <math>AB</math> and <math>BC</math> respectively. Then <math>BP + CQ = BC</math> if and only if <math>APIQ</math> is a | Lemma. Let <math>I</math> be the in-center of <math>ABC</math> and points <math>P</math> and <math>Q</math> be on the lines <math>AB</math> and <math>BC</math> respectively. Then <math>BP + CQ = BC</math> if and only if <math>APIQ</math> is a | ||
− | Solution. Assume that rays <math>AD</math> and <math>BC</math> intersect at point <math>P</math>. Let <math>S</math> be the center od circle touching <math>AD</math>, <math>DC</math> and <math>CB</math>. Obviosuly <math>S</math> is a <math>P</math>ex-center of <math>PDB</math> | + | Solution. Assume that rays <math>AD</math> and <math>BC</math> intersect at point <math>P</math>. Let <math>S</math> be the center od circle touching <math>AD</math>, <math>DC</math> and <math>CB</math>. Obviosuly <math>S</math> is a <math>P</math>-ex-center of <math>PDB</math>, hence <math>DSI=DSP = 1/2DCP=1/2A=DAI</math> so DASI is concyclic. |
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Revision as of 10:09, 9 October 2018
Contents
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.
Possible solution, maybe bogus?
The only way for AD and BC to be tangent to circle O and have AB pass through O is if and
are both 90. But since ABCD is cyclic, the other angles must be 90 as well. Now call the point of tangency of CD E, and since AO=EO, AEOD is a square. Similarily, BCEO is a square, too, so DA=AO and CB=BO. Therefore, AD+BC=AB.
Solution 6
Lemma. Let be the in-center of
and points
and
be on the lines
and
respectively. Then
if and only if
is a
Solution. Assume that rays and
intersect at point
. Let
be the center od circle touching
,
and
. Obviosuly
is a
-ex-center of
, hence
so DASI is concyclic.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Observations
Observe by take ,
on
extended and
1985 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |