# Difference between revisions of "2007 BMO Problems/Problem 1"

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− | Let <math><BAC = x</math> and <math><BDC = y</math>. | + | Let <math><BAC = x</math> and <math><BDC = y</math>. Then by the isosceles triangles manifest in the figure we have <math><DBC = y</math> and <math><ACB = x</math>, so <math><BEA = x+y</math> and <math><EAD = <EDA = \frac{x+y}{2}</math>. Furthermore <math><AEB = 180^\circ - 2x - y</math> and <math><DCE = 180^\circ - x - 2y</math>. |

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+ | If <math>AE = DE</math>, then <math>BE \neq CE</math>. But also <math>AB = CD</math>, so by SSA "Incongruence" (aka. the Law of Sines: <math>\frac{AE}{\sin <ABE} = \frac{AB}{\sin <BEA} = \frac{CD}{\sin <CED} = \frac{DE}{\sin <ECD}</math>) we have <math><ABE + <DCE = 180^\circ</math>. This translates into <math>180^\circ = 3x + 3y</math>, or <math>120^\circ = 2x + 2y</math>, which incidentally equals <math><BAD + <ADC</math>, as desired. | ||

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+ | If <math><BAD + <ADC = 120^\circ</math>, then also <math>x + y + <EAD + <EDA = x + y + (x + y) = 120^\circ</math> by the Exterior Angle Theorem, so <math>3x + 3y = 180^\circ</math> and hence <math><ABE</math> and <math><DCE</math> are supplementary. A simple Law of Sines calculation then gives <math>AE = DE</math>, as desired. This completes both directions of the proof. | ||

## Latest revision as of 23:59, 14 September 2014

## Contents

## Problem

(*Albania*)
Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .

## Solution

Since , , and similarly, . Since , by considering triangles we have . It follows that .

Now, by the Law of Sines,

.

It follows that if and only if

.

Since ,

and

From these inequalities, we see that if and only if (i.e., ) or (i.e., ). But if , then triangles are congruent and , a contradiction. Thus we conclude that if and only if , Q.E.D.

## Solution 2

Let and . Then by the isosceles triangles manifest in the figure we have and , so and . Furthermore and .

If , then . But also , so by SSA "Incongruence" (aka. the Law of Sines: ) we have . This translates into , or , which incidentally equals , as desired.

If , then also by the Exterior Angle Theorem, so and hence and are supplementary. A simple Law of Sines calculation then gives , as desired. This completes both directions of the proof.

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