Difference between revisions of "1985 IMO Problems/Problem 1"

Revision as of 12:34, 3 April 2018

Problem

A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$ . The other three sides are tangent to the circle. Prove that $AD + BC = AB$ .

Solutions

Solution 1

Let $O$ be the center of the circle mentioned in the problem. Let $T$ be the second intersection of the circumcircle of $CDO$ with $AB$ . By measures of arcs, $\angle DTA = \angle DCO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2}$ . It follows that $AT = AD$ . Likewise, $TB = BC$ , so $AD + BC = AB$ , as desired.

Solution 2

Let $O$ be the center of the circle mentioned in the problem, and let $T$ be the point on $AB$ such that $AT = AD$ . Then $\angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO$ , so $DCOT$ is a cyclic quadrilateral and $T$ is in fact the $T$ of the previous solution. The conclusion follows.

Solution 3

Let the circle have center $O$ and radius $r$ , and let its points of tangency with $BC, CD, DA$ be $E, F, G$ , respectively. Since $OEFC$ is clearly a cyclic quadrilateral, the angle $COE$ is equal to half the angle $GAO$ . Then

$\begin{matrix} {CE} & = & r \tan(COE) \\ & = &r \left( \frac{1 - \cos (GAO)}{\sin(GAO)} \right) \\ & = & AO - AG \\ \end{matrix}$

Likewise, $DG = OB - EB$ . It follows that

${EB} + CE + DG + GA = AO + OB$ ,

Q.E.D.

Solution 4

We use the notation of the previous solution. Let $X$ be the point on the ray $AD$ such that $AX = AO$ . We note that $OF = OG = r$ ; $\angle OFC = \angle OGX = \frac{\pi}{2}$ ; and $\angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}$ ; hence the triangles $OFC, OGX$ are congruent; hence $GX = FC = CE$ and $AO = AG + GX = AG + CE$ . Similarly, $OB = EB + GD$ . Therefore $AO + OB = AG + GD + CE + EB$ , 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 $\angle{CBA}$ and $\angle{DAB}$ 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.

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 $M$ , $N$ on $AD$ extended and $BC$

1985 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

@@ Line 39: / Line 39: @@
 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.
+===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 <math>\angle{CBA}</math> and <math>\angle{DAB}</math> 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.

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