1998 JBMO Problems/Problem 2

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.

Solutions

Solution 1

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 $Kris17$

Solution 2

Let $BC = x, DE = y$ . Denote the area of $\triangle XYZ$ by $[XYZ]$ .

$[ABC]+[AED]=\frac{1}{2}(x+y)=\frac{1}{2}$

$[ACD]$ can be found by Heron's formula.

$AC=\sqrt{x^2+1}$

$AD=\sqrt{y^2+1}$

Let $AC=b, AD=c$ .

\begin{align*} [ACD]&=\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\ &=\frac{1}{4}\sqrt{((b+c)^2-1)(1-(b-c)^2)}\\ &=\frac{1}{4}\sqrt{(b+c)^2+(b-c)^2-(b^2-c^2)^2-1}\\ &=\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2)^2-1}\\ &=\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2)^2-1}\\ &=\frac{1}{4}\sqrt{2((x+y)^2-2xy+2)-(x+y)^2(x-y)^2-1}\\ &=\frac{1}{4}\sqrt{5-4xy-(x-y)^2}\\ &=\frac{1}{4}\sqrt{5-(x+y)^2}\\ &=\frac{1}{2} \end{align*}

Total area $=[ABC]+[AED]+[ACD]=\frac{1}{2}+\frac{1}{2}=1$

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1998 JBMO Problems/Problem 2

Contents

Problem 2

Solutions

Solution 1

Solution 2