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

(→‎Solution 6)
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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.
  
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===Possible solution, maybe bogus?===
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===Solution 5===
−
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.
+
 +
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
 +
 
 +
 
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<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

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.

Solution 5

From the fact that AD and BC are tangents to the circle mentioned in the problem, we have


$\angle{CBA}=90\deg$

$\angle{DAB}=90\deg$


Now, from the fact that ABCD is cyclic, we obtain that


$\angle{BCD}=90\deg$

$\angle{CDA}=90\deg$


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


$\angle{OEC}=\angle{OED}=90\deg$


And since $AO=EO=BO$, we obtain two squares, AOED and BOEC. From the properties of squares we now have


$AD+BC=AO+BO=AB$


as desired.

Solution 6

Lemma. Let $I$ be the in-center of $ABC$ and points $P$ and $Q$ be on the lines $AB$ and $BC$ respectively. Then $BP + CQ = BC$ if and only if $APIQ$ is a cyclic quadrilateral.

Solution. Assume that rays $AD$ and $BC$ intersect at point $P$. Let $S$ be the center od circle touching $AD$, $DC$ and $CB$. Obviosuly $S$ is a $P$-ex-center of $PDB$, hence $\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI$ 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 $M$, $N$ 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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