Difference between revisions of "1985 IMO Problems/Problem 1"
Nitinjan06 (talk | contribs) (Incorrect solution flagged.) |
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===Solution 5=== | ===Solution 5=== | ||
− | This solution is incorrect. The fact that < | + | This solution is incorrect. The fact that <math>BC</math> is tangent to the circle does not necessitate that <math>B</math> is its point of tangency. -Nitinjan06 |
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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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>\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI</math> so DASI is concyclic. | 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>\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI</math> so DASI is concyclic. | ||
+ | == Video Solution == | ||
+ | === Solution 1 === | ||
+ | https://www.youtube.com/watch?v=tM0WhXNCWGU | ||
− | + | === Solution 2 === | |
− | + | https://youtu.be/Ormv0y4ZM1E | |
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+ | {{alternate solutions}} | ||
{{IMO box|year=1985|before=First question|num-a=2}} | {{IMO box|year=1985|before=First question|num-a=2}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 07:28, 3 October 2021
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
This solution is incorrect. The fact that is tangent to the circle does not necessitate that
is its point of tangency. -Nitinjan06
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.
Video Solution
Solution 1
https://www.youtube.com/watch?v=tM0WhXNCWGU
Solution 2
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 |