Difference between revisions of "1985 IMO Problems/Problem 1"
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We use the notation of the previous solution. Let <math>X</math> be the point on the ray <math>AD</math> such that <math>AX = AO</math>. We note that <math>OF = OG = r </math>; <math> \angle OFC = \angle OGX = \frac{\pi}{2} </math>; and <math> \angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}</math>; hence the triangles <math>OFC, OGX</math> are congruent; hence <math>GX = FC = CE </math> and <math>AO = AG + GX = AG + CE</math>. Similarly, <math>OB = EB + GD </math>. Therefore <math>AO + OB = AG + GD + CE + EB </math>, Q.E.D. | We use the notation of the previous solution. Let <math>X</math> be the point on the ray <math>AD</math> such that <math>AX = AO</math>. We note that <math>OF = OG = r </math>; <math> \angle OFC = \angle OGX = \frac{\pi}{2} </math>; and <math> \angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}</math>; hence the triangles <math>OFC, OGX</math> are congruent; hence <math>GX = FC = CE </math> and <math>AO = AG + GX = AG + CE</math>. Similarly, <math>OB = EB + GD </math>. Therefore <math>AO + OB = AG + GD + CE + EB </math>, Q.E.D. | ||
− | === | + | ===Solution 5=== |
− | + | ||
+ | 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{DAB}=90\deg</math> | ||
+ | |||
+ | |||
+ | Now, from the fact that ABCD is cyclic, we obtain that | ||
+ | |||
+ | |||
+ | <math>\angle{BCD}=90\deg</math> | ||
+ | |||
+ | <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 | ||
+ | |||
+ | |||
+ | <math>\angle{OEC}=\angle{OED}=90\deg</math> | ||
+ | |||
+ | |||
+ | And since <math>AO=EO=BO</math>, we obtain two squares, AOED and BOEC. | ||
+ | From the properties of squares we now have | ||
+ | |||
+ | |||
+ | <math>AD+BC=AO+BO=AB</math> | ||
+ | |||
+ | |||
+ | as desired. | ||
=== Solution 6 === | === Solution 6 === |
Revision as of 05:59, 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
Now, from the fact that ABCD is cyclic, we obtain that
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
And since , we obtain two squares, AOED and BOEC.
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 |