Difference between revisions of "2012 AMC 8 Problems/Problem 25"

Latest revision as of 01:59, 25 December 2024

Problem

A square with area $4$ is inscribed in a square with area $5$ , with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$ , and the other of length $b$ . What is the value of $ab$ ?

$[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]$

$\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$

Solution 1

The total area of the four congruent triangles formed by the squares is $5-4 = 1$ . Therefore, the area of one of these triangles is $\frac{1}{4}$ . The height of one of these triangles is $a$ and the base is $b$ . Using the formula for area of the triangle, we have $\frac{ab}{2} = \frac{1}{4}$ . Multiply by $2$ on both sides to find that the value of $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$ .

Solution 2 (Algebra)

We see that we want $ab$ , so instead of solving for $a,b$ , we find a way to get an expression with $ab$ .

By Triple Perpendicularity Model,

all four triangles are congruent.

By Pythagorean's Theorem,

$\sqrt{a^2+b^2} = \sqrt{4}$

Thus, $\sqrt{a^2+b^2} = 2$

As $a+b=\sqrt{5}$ ,

$a^2+2ab+b^2 = 5$

So, $\sqrt{5-2ab} = 2$

Simplifying,

$5-2ab = 4$

$-2ab=-1$

$ab=\frac{1}{2}$ or $\boxed{\textbf{(C)} \frac{1}{2}}$

~ lovelearning999

Solution 3 (similar to solution 2)

We know that each side of a square is equal, and each the area of a square can be expressed as the side squared. We can let the outside square with area 5's side be $x$ . We get the equation $x^2 = 5$ . Simplifying this we get $x=\sqrt5$ .

We can then create the equation $a+b=\sqrt5$ .

Using the same tactic we get that the side length of the inner square is $2$ . By the Pythagorean Theorem,

$a^2 + b^2 = 2^2$ .

We can then express this expression as

$(a+b)^2 - 2ab = 4$ .

We recall that $a+b=\sqrt5$ and substitute it into our current equation:

$(\sqrt5)^2 - 2ab = 4$ .

We further simplify this and end up with,

$ab=1/2$ which is $\boxed{\textbf{(C)} \frac {1}{2}}$

Solution 4 (The Long Way)

Visually, $a+b=\sqrt5$ (1) and we can tell that all the triangles are congruent. That means the long side of all the triangles is $b$ and the short side will always be $a$ . Notice that the hypotenuse is the side of the smaller square (2). Using the Pythagorean theorem, $a^2+b^2=2^2$ or $a^2+b^2=4$ (2).

Now we just have a system of equations:

Isolate b in (1) -> $b=\sqrt5-a$ , and plug it in to (2).

$a^2+(5-a)^2=4$ , Simplifying should get you to $2a^2-2a\sqrt5+1=0$

Using the quadratic formula, $a=\frac{2\sqrt5 \pm \sqrt{20-8}}{4}$

Simplifying the squareroots, $a=\frac{\sqrt5+\sqrt3}{2}$ .

This yields b to be $\frac{\sqrt5-\sqrt3}{2}$ via substituting into (1)

$ab=\frac{(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)}{2*2}$ (difference of squares) $= \frac{5-3}{4} = \boxed{\textbf{(C)} \frac {1}{2}}$

~RandomMathGuy500

Video Solution 2

https://youtu.be/MhxGq1sSA6U ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2

~ pi_is_3.14

@@ Line 40: / Line 40: @@
 <math>ab=\frac{1}{2}</math> or <math>\boxed{\textbf{(C)} \frac{1}{2}}</math>
+~ lovelearning999
+==Solution 3 (similar to solution 2)==
+We know that each side of a square is equal, and each the area of a square can be expressed as the side squared. We can let the outside square with area 5's side be <math>x</math>. We get the equation <math>x^2 = 5</math>. Simplifying this we get <math>x=\sqrt5</math>.
+We can then create the equation <math>a+b=\sqrt5</math>.
+Using the same tactic we get that the side length of the inner square is <math>2</math>. By the Pythagorean Theorem,
+<math>a^2 + b^2 = 2^2</math>.
+We can then express this expression as
+<math>(a+b)^2 - 2ab = 4</math>.
+We recall that <math>a+b=\sqrt5</math> and substitute it into our current equation:
+<math>(\sqrt5)^2 - 2ab = 4</math>.
+We further simplify this and end up with,
+<math>ab=1/2</math> which is <math>\boxed{\textbf{(C)} \frac {1}{2}}</math>
+==Solution 4 (The Long Way)==
+Visually, <math>a+b=\sqrt5</math> (1) and we can tell that all the triangles are congruent. That means the long side of all the triangles is <math>b</math> and the short side will always be <math>a</math>. Notice that the hypotenuse is the side of the smaller square (2). Using the Pythagorean theorem, <math>a^2+b^2=2^2</math> or <math>a^2+b^2=4</math> (2).
+Now we just have a system of equations:
+Isolate b in (1) -> <math>b=\sqrt5-a</math>, and plug it in to (2).
+<math>a^2+(5-a)^2=4</math>, Simplifying should get you to <math>2a^2-2a\sqrt5+1=0</math>
+Using the quadratic formula, <math>a=\frac{2\sqrt5 \pm \sqrt{20-8}}{4}</math>
+Simplifying the squareroots, <math>a=\frac{\sqrt5+\sqrt3}{2}</math>.
+This yields b to be <math>\frac{\sqrt5-\sqrt3}{2}</math> via substituting into (1)
+<math>ab=\frac{(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)}{2*2}</math> (difference of squares) <math>= \frac{5-3}{4} = \boxed{\textbf{(C)} \frac {1}{2}}</math>
+~RandomMathGuy500
 ==Video Solution 2==
 https://youtu.be/MhxGq1sSA6U ~savannahsolver

2012 AMC 8 (Problems • Answer Key • Resources)
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