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Difference between revisions of "1985 IMO Problems/Problem 1"
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Lemma 2. Let <math>I</math> be the in-center of <math>APB</math>. Then <math>ID = IC = IL</math> where <math>L</math> is the intersection of <math>PI</math> and <math>AB</math>.
 
Lemma 2. Let <math>I</math> be the in-center of <math>APB</math>. Then <math>ID = IC = IL</math> where <math>L</math> is the intersection of <math>PI</math> and <math>AB</math>.
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Revision as of 10:38, 1 October 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.

Solution 6

Let P be intersection of rays $AD$ and $BC$.

Lemma 1. A circle passes through points $A$, any points $D$, $C$ on triangle $APB$ and the in-center of $APB$ if and only if $AD + BC = AB$.

Lemma 2. Let $I$ be the in-center of $APB$. Then $ID = IC = IL$ where $L$ is the intersection of $PI$ and $AB$.



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
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