# Difference between revisions of "1984 IMO Problems/Problem 4"

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+ | ==Problem== | ||

Let <math>ABCD</math> be a convex quadrilateral with the line <math>CD</math> being tangent to the circle on diameter <math>AB</math>. Prove that the line <math>AB</math> is tangent to the circle on diameter <math>CD</math> if and only if the lines <math>BC</math> and <math>AD</math> are parallel. | Let <math>ABCD</math> be a convex quadrilateral with the line <math>CD</math> being tangent to the circle on diameter <math>AB</math>. Prove that the line <math>AB</math> is tangent to the circle on diameter <math>CD</math> if and only if the lines <math>BC</math> and <math>AD</math> are parallel. | ||

+ | ==Solution== | ||

First, we prove that if <math>BC</math> and <math>AD</math> are parallel then the claim is true: Let <math>AB</math> and <math>CD</math> intersect at <math>E</math> (assume <math>E</math> is closer to <math>AD</math>, the other case being analogous). Let <math>M,N</math> be the midpoints of <math>AB,CD</math> respectively. Let the length of the perpendicular from <math>N</math> to <math>AB</math> be <math>r</math>. It is known that the length of the perpendicular from <math>M</math> to <math>CD</math> is <math>\frac{1}{2}AB</math>. Let the foot of the perpendicular from <math>C</math> to <math>AB</math> be <math>H</math>, and similarly define <math>G</math> for side <math>CD</math>. Then, since triangles <math>MNE</math> and <math>BCE</math> are similar, we have <math>\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}</math>. This gives an expression for <math>r</math>: | First, we prove that if <math>BC</math> and <math>AD</math> are parallel then the claim is true: Let <math>AB</math> and <math>CD</math> intersect at <math>E</math> (assume <math>E</math> is closer to <math>AD</math>, the other case being analogous). Let <math>M,N</math> be the midpoints of <math>AB,CD</math> respectively. Let the length of the perpendicular from <math>N</math> to <math>AB</math> be <math>r</math>. It is known that the length of the perpendicular from <math>M</math> to <math>CD</math> is <math>\frac{1}{2}AB</math>. Let the foot of the perpendicular from <math>C</math> to <math>AB</math> be <math>H</math>, and similarly define <math>G</math> for side <math>CD</math>. Then, since triangles <math>MNE</math> and <math>BCE</math> are similar, we have <math>\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}</math>. This gives an expression for <math>r</math>: | ||

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Since <math>\frac{1}{2}CD\ne EN</math>, we have <math>NT=0</math> as desired. | Since <math>\frac{1}{2}CD\ne EN</math>, we have <math>NT=0</math> as desired. | ||

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+ | ==See also== | ||

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+ | [[Category:Olympiad Geometry Problems]] |

## Revision as of 10:26, 25 August 2008

## Problem

Let be a convex quadrilateral with the line being tangent to the circle on diameter . Prove that the line is tangent to the circle on diameter if and only if the lines and are parallel.

## Solution

First, we prove that if and are parallel then the claim is true: Let and intersect at (assume is closer to , the other case being analogous). Let be the midpoints of respectively. Let the length of the perpendicular from to be . It is known that the length of the perpendicular from to is . Let the foot of the perpendicular from to be , and similarly define for side . Then, since triangles and are similar, we have . This gives an expression for :

Noticing that simplifies the expression to

By the Law of Sines, . Since triangles are similar, we have and thus we have

and we are done.

Now to prove the converse. Suppose we have the quadrilateral with parallel to , and with all conditions satisfied. We shall prove that there exists no point on such that is a midpoint of a side of a quadrilateral which also satisfies the condition. Suppose there was such a . Like before, define the points for quadrilateral . Let be the length of the perpendicular from to . Then, using similar triangles, . This gives

But, we must have . Thus, we have

Since , we have as desired.