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Difference between revisions of "1998 JBMO Problems/Problem 2"
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==Problem 2==
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Let <math>ABCDE</math> be a convex pentagon such that <math>AB=AE=CD=1</math>, <math>\angle ABC=\angle DEA=90^\circ</math> and <math>BC+DE=1</math>. Compute the area of the pentagon.
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== Solution ==
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Let <math>BC = a, ED = 1 - a</math>
 
Let <math>BC = a, ED = 1 - a</math>
  

Revision as of 23:18, 3 December 2018

Problem 2

Let $ABCDE$ be a convex pentagon such that $AB=AE=CD=1$, $\angle ABC=\angle DEA=90^\circ$ and $BC+DE=1$. Compute the area of the pentagon.


Solution

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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