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

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== Problem ==
 
== Problem ==
  
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A [[circle]] has center on the side <math>\displaystyle AB</math> of the [[cyclic quadrilateral]] <math>\displaystyle ABCD</math>.  The other three sides are [[tangent]] to the circle.  Prove that <math>\displaystyle AD + BC = AB</math>.
+
A [[circle]] has center on the side <math>AB</math> of the [[cyclic quadrilateral]] <math>ABCD</math>.  The other three sides are [[tangent]] to the circle.  Prove that <math>AD + BC = AB</math>.
  
 
== Solutions ==
 
== Solutions ==
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=== Solution 2 ===
 
=== Solution 2 ===
  
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Let <math> \displaystyle O </math> be the center of the circle mentioned in the problem, and let <math>\displaystyle T</math> be the point on <math>\displaystyle AB </math> such that <math> \displaystyle AT = AD </math>.  Then <math> \displaystyle \angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO</math>, so <math> \displaystyle DCOT </math> is a cyclic quadrilateral and <math> \displaystyle T </math> is in fact the <math> \displaystyle T</math> of the previous solution.  The conclusion follows.
+
Let <math>O </math> be the center of the circle mentioned in the problem, and let <math>T</math> be the point on <math>AB </math> such that <math>AT = AD </math>.  Then <math>\angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO</math>, so <math>DCOT </math> is a cyclic quadrilateral and <math>T </math> is in fact the <math>T</math> of the previous solution.  The conclusion follows.
  
 
=== Solution 3 ===
 
=== Solution 3 ===
  
−
Let the circle have center <math> \displaystyle O </math> and radius <math> \displaystyle r </math>, and let its points of tangency with <math> \displaystyle BC, CD, DA </math> be <math> \displaystyle E, F, G </math>, respectively.  Since <math> \displaystyle OEFC </math> is clearly a cyclic quadrilateral, the angle <math> \displaystyle COE </math> is equal to half the angle <math> \displaystyle GAO </math>.  Then
+
Let the circle have center <math>O </math> and radius <math>r </math>, and let its points of tangency with <math>BC, CD, DA </math> be <math>E, F, G </math>, respectively.  Since <math>OEFC </math> is clearly a cyclic quadrilateral, the angle <math>COE </math> is equal to half the angle <math>GAO </math>.  Then
  
 
<center>
 
<center>
 
<math>
 
<math>
 
\begin{matrix} {CE} & = & r \tan(COE) \\
 
\begin{matrix} {CE} & = & r \tan(COE) \\
−
& = & \displaystyle r \left( \frac{1 - \cos (GAO)}{\sin(GAO)} \right) \\
+
& = &r \left( \frac{1 - \cos (GAO)}{\sin(GAO)} \right) \\
 
& = & AO - AG \\
 
& = & AO - AG \\
 
\end{matrix}
 
\end{matrix}
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</center>
 
</center>
  
−
Likewise, <math>\displaystyle DG = OB - EB</math>.  It follows that
+
Likewise, <math>DG = OB - EB</math>.  It follows that
  
 
<center>
 
<center>
 
<math>
 
<math>
−
\displaystyle {EB} + CE + DG + GA = AO + OB
+
{EB} + CE + DG + GA = AO + OB
 
</math>,
 
</math>,
 
</center>
 
</center>
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=== Solution 4 ===
 
=== Solution 4 ===
  
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We use the notation of the previous solution.  Let <math>\displaystyle X</math> be the point on the ray <math>\displaystyle AD</math> such that <math> \displaystyle AX = AO</math>.  We note that <math>\displaystyle 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>\displaystyle OFC, OGX</math> are congruent; hence <math> \displaystyle GX = FC = CE </math> and <math> \displaystyle AO = AG + GX = AG + CE</math>.  Similarly, <math> \displaystyle OB = EB + GD </math>.  Therefore <math> \displaystyle 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.
 +
 
 +
===Solution 5===
 +
 +
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
 +
<math>\angle{CBA}=90\deg</math>
 +
and
 +
<math>\angle{DAB}=90\deg</math>.
 +
 
 +
Now, from the fact that ABCD is cyclic, we obtain that
 +
<math>\angle{BCD}=90\deg</math>
 +
and
 +
<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
 +
<math>\angle{OEC}=\angle{OED}=90\deg</math>
 +
 
 +
Since <math>AO=EO=BO</math>, we obtain two squares, <math>AOED</math> and <math>BOEC</math>.
 +
From the properties of squares we now have
 +
 
 +
 
 +
<math>AD+BC=AO+BO=AB</math>
 +
 
 +
 
 +
as desired.
 +
 
 +
=== 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 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>, hence <math>\angle DSI=\angle DSP = \frac{1}{2} \angle DCP=\frac{1}{2} \angle A=\angle DAI</math> so DASI is concyclic.
 +
 
 +
== Video Solution ==
 +
 
 +
=== Solution 1 ===
 +
https://www.youtube.com/watch?v=tM0WhXNCWGU
 +
 
 +
 
 +
=== Solution 2 ===
 +
https://youtu.be/Ormv0y4ZM1E
  
  
 
{{alternate solutions}}
 
{{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
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Problem 2
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