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

Latest revision as of 08:31, 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 quadrilateral $AD'CD$ .

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 4: / Line 4: @@
-== Solution ==
+== Solutions ==
+=== Solution 1 ===
 Let <math>BC = a, ED = 1 - a</math>
-Let angle <math>DAC</math> = <math>X</math>
+Let <math>\angle DAC = X</math>
-Applying cosine rule to triangle <math>DAC</math> we get:
+Applying cosine rule to <math>\triangle DAC</math> we get:
-<math>Cos X = (AC ^ {2} + AD ^ {2} - DC ^ {2}) / (2 * AC * AD )</math>
+<math>\cos X = \frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \cdot AC \cdot AD }</math>
 Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get:
-<math>Cos^{2} X = (1 - a - a ^ {2}) ^ {2} / ((1 + a^{2})(2 - 2a + a^{2}))</math>
+<math>\cos^{2} X = \frac{(1 - a - a ^ {2}) ^ {2}}{(1 + a^{2})(2 - 2a + a^{2})}</math>
-From above, <math>Sin^{2} X = 1 - Cos^{2} X  =  1 / ((1 + a^{2})(2 - 2a + a^{2})) = 1/(AC^{2}.AD^{2})</math>
+From above, <math>\sin^{2} X = 1 - \cos^{2} X  =  \frac{1}{(1 + a^{2})(2 - 2a + a^{2})} = \frac{1}{AC^{2} \cdot AD^{2}}</math>
-Thus,  <math>Sin X * AC * AD = 1</math>
+Thus,  <math>\sin X \cdot AC \cdot AD = 1</math>
-So, <math>Area</math> of triangle <math>DAC</math> = <math>(1/2)*Sin X * AC * AD = 1/2</math>
+So, area of <math>\triangle DAC</math> = <math>\frac{1}{2}\cdot \sin X \cdot AC \cdot AD = \frac{1}{2}</math>
-Let <math>AF</math> be the altitude of triangle DAC from A.
+Let <math>AF</math> be the altitude of <math>\triangle DAC</math> from <math>A</math>.
-So <math>1/2*DC*AF = 1/2</math>
+So <math>\frac{1}{2}\cdot DC\cdot AF = \frac{1}{2}</math>
 This implies <math>AF = 1</math>.
-Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, traingle <math>ABC</math> is congruent to <math>AFC</math>.
+Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, <math>\triangle ABC</math> is congruent to <math>\triangle AFC</math>.
-Similarly <math>AEDF</math> is a cyclic quadrilateral and traingle <math>AED</math> is congruent to <math>AFD</math>.
+Similarly <math>AEDF</math> is a cyclic quadrilateral and <math>\triangle AED</math> is congruent to <math>\triangle AFD</math>.
-So <math>area</math> of triangle <math>ABC</math> + <math>area</math> of triangle <math>AED</math> = <math>area</math> of Triangle <math>ADC</math>.
+So area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> = area of <math>\triangle ADC</math>.
-Thus <math>area</math> of pentagon <math>ABCD</math> = <math>area</math> of <math>ABC</math> + <math>area</math> of <math>AED</math> + <math>area</math> of <math>DAC</math> = <math>1/2 + 1/2 = 1</math>
+Thus area of pentagon <math>ABCD</math> = area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> + area of <math>\triangle DAC</math> = <math>\frac{1}{2}+\frac{1}{2} = 1</math>
 By <math>Kris17</math>
+=== Solution 2 ===
+Let <math>BC = x, DE = y</math>. Denote the area of <math>\triangle XYZ</math> by <math>[XYZ]</math>.
+<math>[ABC]+[AED]=\frac{1}{2}(x+y)=\frac{1}{2}</math>
+<math>[ACD]</math> can be found by [[Heron's formula]].
+<math>AC=\sqrt{x^2+1}</math>
+<math>AD=\sqrt{y^2+1}</math>
+Let <math>AC=b, AD=c</math>.
+<cmath>
+\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*}
+</cmath>
+Total area <math>=[ABC]+[AED]+[ACD]=\frac{1}{2}+\frac{1}{2}=1</math>.
+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 quadrilateral <math>AD'CD</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 <math>[ACD]</math> = <math>\frac{1}{2}</math>
+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
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