Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (→‎Solution 1)
Line 18: Line 18:
 
Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get:
 
Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get:
  
−
<math>\cos^{2} X = \frac{(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  =  \frac{1}{((1 + a^{2})(2 - 2a + a^{2}))} = \frac{1}{AC^{2} \cdot 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 \cdot AC \cdot AD = 1</math>
 
Thus,  <math>\sin X \cdot AC \cdot AD = 1</math>

Revision as of 23:26, 4 June 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_JBMO_Problems/Problem_2&oldid=123853"