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

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===Solution 5===
 
===Solution 5===
 
   
 
   
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This solution is incorrect. The fact that <cmath>BC</cmath> is tangent to the circle does not necessitate that <cmath>B</cmath> is its point of tangency.
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This solution is incorrect. The fact that <math>BC</math> is tangent to the circle does not necessitate that <math>B</math> is its point of tangency. -Nitinjan06
  
 
From the fact that AD and BC are tangents to the circle mentioned in the problem, we have
 
From the fact that AD and BC are tangents to the circle mentioned in the problem, we have
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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>, hence <math>\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI</math> so DASI is concyclic.
 
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>, hence <math>\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI</math> so DASI is concyclic.
  
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== Video Solution ==
  
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=== Solution 1 ===
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https://www.youtube.com/watch?v=tM0WhXNCWGU
  
  
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{{alternate solutions}}
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=== Solution 2 ===
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https://youtu.be/Ormv0y4ZM1E
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Observations
 
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Observe by take <math>M</math>, <math>N</math> on <math>AD</math> extended and
 
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<math>BC</math>
 
  
  
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{{alternate solutions}}
  
 
{{IMO box|year=1985|before=First question|num-a=2}}
 
{{IMO box|year=1985|before=First question|num-a=2}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 06:28, 3 October 2021

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

This solution is incorrect. The fact that $BC$ is tangent to the circle does not necessitate that $B$ is its point of tangency. -Nitinjan06

From the fact that AD and BC are tangents to the circle mentioned in the problem, we have $\angle{CBA}=90\deg$ and $\angle{DAB}=90\deg$.

Now, from the fact that ABCD is cyclic, we obtain that $\angle{BCD}=90\deg$ and $\angle{CDA}=90\deg$, 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 $\angle{OEC}=\angle{OED}=90\deg$

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.

Video Solution

Solution 1

https://www.youtube.com/watch?v=tM0WhXNCWGU


Solution 2

https://youtu.be/Ormv0y4ZM1E


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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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