Difference between revisions of "1985 IMO Problems/Problem 1"
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=== Solution 6 === | === Solution 6 === | ||
− | 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>. By angle-chasing show that <math>PDIC</math> is a cyclic quadrilateral. | + | 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>. By angle-chasing show that <math>PDIC</math> is a cyclic quadrilateral. |
Revision as of 09:04, 9 October 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.
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 . By angle-chasing show that is a cyclic quadrilateral.
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 |