Difference between revisions of "1985 IMO Problems/Problem 1"
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From the fact that AD and BC are tangents to the circle mentioned in the problem, we have | From the fact that AD and BC are tangents to the circle mentioned in the problem, we have | ||
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<math>\angle{CBA}=90\deg</math> | <math>\angle{CBA}=90\deg</math> | ||
− | + | and | |
− | <math>\angle{DAB}=90\deg</math> | + | <math>\angle{DAB}=90\deg</math>. |
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Now, from the fact that ABCD is cyclic, we obtain that | Now, from the fact that ABCD is cyclic, we obtain that | ||
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<math>\angle{BCD}=90\deg</math> | <math>\angle{BCD}=90\deg</math> | ||
+ | and | ||
+ | <math>\angle{CDA}=90\deg</math>, | ||
+ | such that ABCD is a rectangle. | ||
− | + | Now, let E be the point of tangency between the circle and CD. It follows, if O is the center of the circle, that | |
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− | Now, let E be the point of tangency between the circle and CD. | ||
− | It follows, if O is the center of the circle, that | ||
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<math>\angle{OEC}=\angle{OED}=90\deg</math> | <math>\angle{OEC}=\angle{OED}=90\deg</math> | ||
− | + | Since <math>AO=EO=BO</math>, we obtain two squares, <math>AOED</math> and <math>BOEC</math>. | |
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From the properties of squares we now have | From the properties of squares we now have | ||
Revision as of 07:09, 26 December 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.
Solution 5
From the fact that AD and BC are tangents to the circle mentioned in the problem, we have
and
.
Now, from the fact that ABCD is cyclic, we obtain that
and
,
such that ABCD is a rectangle.
Now, let E be the point of tangency between the circle and CD. It follows, if O is the center of the circle, that
Since , we obtain two squares,
and
.
From the properties of squares we now have
as desired.
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 cyclic quadrilateral.
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 |