Difference between revisions of "1998 JBMO Problems/Problem 2"

Revision as of 07:29, 2 July 2020

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:

$\cos X = \frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \cdot AC \cdot AD }$

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

$\cos^{2} X = \frac{(1 - a - a ^ {2}) ^ {2}}{(1 + a^{2})(2 - 2a + a^{2})}$

From above, $\sin^{2} X = 1 - \cos^{2} X = \frac{1}{(1 + a^{2})(2 - 2a + a^{2})} = \frac{1}{AC^{2} \cdot AD^{2}}$

Thus, $\sin X \cdot AC \cdot AD = 1$

So, area of $\triangle DAC$ = $\frac{1}{2}\cdot \sin X \cdot AC \cdot AD = \frac{1}{2}$

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

So $\frac{1}{2}\cdot DC\cdot AF = \frac{1}{2}$

This implies $AF = 1$ .

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

So area of $\triangle ABC$ + area of $\triangle AED$ = area of $\triangle ADC$ . Thus area of pentagon $ABCD$ = area of $\triangle ABC$ + area of $\triangle AED$ + area of $\triangle DAC$ = $\frac{1}{2}+\frac{1}{2} = 1$

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

By durianice

Solution 3

Construct $AD$ and $AC$ to partition the figure into $ABC$ , $ACD$ and $ADE$ .

Rotate $ADE$ with centre $A$ such that $AE$ coincides with $AB$ and $AD$ is mapped to $AD'$ . Hence the area of the pentagon is still preserved and it suffices to find the area of the pentagon $AD'BCD$ .

Hence $[AD'C]$ = $\frac{1}{2}$ ( $D'E + BC$ ) $AB$ = $\frac{1}{2}$

Since $CD$ = $CD'$ , $AC$ = $AC$ and $AD$ = $AD'$ , by SSS Congruence, $ACD$ and $ACD'$ are congruent, so [ACD]&=\frac{1}{2}\

So the area of pentagon $ABCDE = \frac{1}{2} + \frac{1}{2} = 1$ .

- SomebodyYouUsedToKnow

@@ Line 73: / Line 73: @@
 By durianice
+=== Solution 3 ===
+Construct <math>AD</math> and <math>AC</math> to partition the figure into <math>ABC</math>, <math>ACD</math> and <math>ADE</math>.
+Rotate <math>ADE</math> with centre <math>A</math> such that <math>AE</math> coincides with <math>AB</math> and <math>AD</math> is mapped to <math>AD'</math>. Hence the area of the pentagon is still preserved and it suffices to find the area of the pentagon <math>AD'BCD</math>.
+Hence <math>[AD'C]</math> = <math>\frac{1}{2}</math> (<math>D'E + BC</math>)<math>AB</math>= <math>\frac{1}{2}</math>
+Since <math>CD</math> = <math>CD'</math>, <math>AC</math> = <math>AC</math> and <math>AD</math> = <math>AD'</math>, by SSS Congruence, <math>ACD</math> and <math>ACD'</math> are congruent, so [ACD]&=\frac{1}{2}\
+So the area of pentagon <math>ABCDE = \frac{1}{2} + \frac{1}{2} = 1</math>.
+- SomebodyYouUsedToKnow
 ==See Also==
 {{JBMO box|year=1998|num-b=1|num-a=3|five=}}

1998 JBMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4
All JBMO Problems and Solutions

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