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

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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. | ||

+ | |||

+ | ==Solution 2== | ||

+ | Let <math>M</math> and <math>N</math> be midpoints of <math>AB</math> and <math>CD</math>, respectively. Let <math>[X]</math> denote the area of figure <math>X</math>. | ||

+ | |||

+ | If <math>AD // BC</math> and the circle centered at <math>M</math> with diameter <math>AB</math> touches <math>CD</math> at <math>T</math>, then <math>MN // AD</math>. It follows that <math>[AMN] = [DMN]</math>, so <math>\frac{1}{2} AM \cdot d = \frac{1}{2} MT \cdot DN</math>, where <math>d</math> is the distance from <math>N</math> to <math>AB</math>. But we have <math>MT = MA</math>, so <math>DN = d</math>. It follows that the circle centered at <math>N</math> with diameter <math>CD</math> touches <math>AB</math>. | ||

+ | |||

+ | If on the other hand we have the circle with diameter <math>CD</math> touching <math>AB</math> at <math>U</math>(as well as the circle with diameter <math>AB</math> touching <math>CD</math> at <math>T</math>), we must have <math>\frac{1}{2} DN \cdot MT = \frac{1}{2} AM \cdot UN</math> because of the equivalence of radii in circles. Hence <math>[ANM] = [DMN]</math>, so <math>A</math> and <math>D</math> are equidistant from <math>MN</math> (as <math>MN</math> is the common base). Hence <math>AB // MN</math>. Similarly <math>CD // MN</math>, and so <math>AB // CD</math>, as desired. | ||

==See also== | ==See also== | ||

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

## Revision as of 23:32, 14 September 2014

## Contents

## 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.

## Solution 2

Let and be midpoints of and , respectively. Let denote the area of figure .

If and the circle centered at with diameter touches at , then . It follows that , so , where is the distance from to . But we have , so . It follows that the circle centered at with diameter touches .

If on the other hand we have the circle with diameter touching at (as well as the circle with diameter touching at ), we must have because of the equivalence of radii in circles. Hence , so and are equidistant from (as is the common base). Hence . Similarly , and so , as desired.