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1998 JBMO Problems/Problem 2

Revision as of 00:19, 29 November 2018 by KRIS17 (talk | contribs)

Let $BC = a, ED = 1 - a$

Let angle $DAC$ = $X$

Applying cosine rule to triangle $DAC$ we get:

$Cos X = (AC ^ {2} + AD ^ {2} - DC ^ {2}) / (2 * AC * AD )$

Substituting $AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1$ we get:

$Cos^{2} X = (1 - a - a ^ {2}) ^ {2} / ((1 + a^{2})(2 - 2a + a^{2}))$

From above, $Sin^{2} X = 1 - Cos^{2} X = 1 / ((1 + a^{2})(2 - 2a + a^{2})) = 1/(AC^{2}.AD^{2})$

Thus, $Sin X * AC * AD = 1$

So, $Area$ of triangle $DAC$ = $(1/2)*Sin X * AC * AD = 1/2$

Let $AF$ be the altitude of triangle DAC from A.

So $1/2*DC*AF = 1/2$

This implies $AF = 1$.

Since $AFCB$ is a cyclic quadrilateral with $AB = AF$, traingle $ABC$ is congruent to $AFC$. Similarly $AEDF$ is a cyclic quadrilateral and traingle $AED$ is congruent to $AFD$.

So $area$ of triangle $ABC$ + $area$ of triangle $AED$ = $area$ of Triangle $ADC$. Thus $area$ of pentagon $ABCD$ = $area$ of $ABC$ + $area$ of $AED$ + $area$ of $DAC$ = $1/2 + 1/2 = 1$

By Kris17

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