Difference between revisions of "1985 IMO Problems/Problem 1"
(→Solution 5) |
|||
Line 43: | Line 43: | ||
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 | ||
− | |||
− | |||
<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>. |
− | |||
Now, from the fact that ABCD is cyclic, we obtain that | Now, from the fact that ABCD is cyclic, we obtain that | ||
− | |||
− | |||
<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 | |
− | |||
− | |||
− | |||
− | Now, let E be the point of tangency between the circle and CD. | ||
− | It follows, if O is the center of the circle, that | ||
− | |||
− | |||
<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>. | |
− | |||
From the properties of squares we now have | From the properties of squares we now have | ||
Revision as of 06:09, 26 December 2018
Contents
[hide]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 |