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

Latest revision as of 19:37, 5 July 2025

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 - Recommended for Contest

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 $a$ and $b$ , we find a way to get an expression with $ab$ .

By the Triple Perpendicularity Model, all four triangles are congruent.

By the Pythagorean 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 to $ab=1/2$ 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

Solution 3 (Pythagorean Theorem & Vieta's Formula)

Proceed as in solution 3 until you get the equation $2a^2-2a\sqrt5+1=0$ . The two solutions to this equation will be the long and short side of the triangle we are looking for, or $a$ and $b$ from the problem statement. Using Vieta's formula, their product is the constant term divided by the quadratic term, or $\boxed{\textbf{(C)} \frac {1}{2}}$ .

~nbeaudrot

Video Solution by OmegaLearn

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

Video Solution 2

https://youtu.be/MhxGq1sSA6U ~savannahsolver

~ pi_is_3.14

@@ Line 1: / Line 1: @@
 ==Problem==
-A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length <math> a </math>, and the other of length <math> b </math>. What is the value of <math> ab </math>?
+A square with area <math>4</math> is inscribed in a square with area <math>5</math>, 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 <math> a </math>, and the other of length <math> b </math>. What is the value of <math> ab </math>?
 <asy>
@@ Line 10: / Line 10: @@
 <math> \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 </math>
-==Solution 1==
+==Solution 1 - Recommended for Contest==
-The total area of the four congruent triangles formed by the squares is <math>5-4 = 1 </math>. Therefore, the area of one of these triangles is  <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> a \cdot b </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
+The total area of the four congruent triangles formed by the squares is <math>5-4 = 1 </math>. Therefore, the area of one of these triangles is  <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
-==Solution 2==
+==Solution 2 (Algebra)==
-To solve this problem you could also use algebraic manipulation.
+We see that we want <math>ab</math>, so instead of solving for <math>a</math> and <math>b</math>, we find a way to get an expression with <math>ab</math>.
-Since the area of the large square is <math> 5 </math>, the sidelength is <math> \sqrt{5} </math>.
+By the Triple Perpendicularity Model, all four triangles are congruent.
-We then have the equation <math> a + b = \sqrt{5} </math>.
+By the Pythagorean Theorem,
-We also know that the side length of the smaller square is  <math> 2 </math>, since its area is <math> 4 </math>. Then, the segment of length <math> a </math> and segment of length <math> b </math> form a right triangle whose hypotenuse would have length <math> 2 </math>.
+<math>\sqrt{a^2+b^2} = \sqrt{4}</math>
-So our second equation is <math> \sqrt{{a^2}+{b^2}} = 2 </math>.
+Thus, <math>\sqrt{a^2+b^2} = 2</math>
-Square both equations.
+As <math>a+b=\sqrt{5}</math>,
-<math> a^2 + 2ab + b^2 = 5 </math>
+<math>a^2+2ab+b^2 = 5</math>
-<math> a^2 + b^2 = 4 </math>
+So, <math>\sqrt{5-2ab} = 2</math>
-Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> a \cdot b </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
+Simplifying,
-==Solution 3 (similar to solution 1)==
+<math>5-2ab = 4</math>
-Since we know 4 of the triangles both have side lengths a and b, we can create an equation, which the area of the inner square plus the sum of the 4 triangles being the area of the outer square.
+<math>-2ab=-1</math>
-<math> 4 + 2ab = 5</math>
+<math>ab=\frac{1}{2}</math> or <math>\boxed{\textbf{(C)} \frac{1}{2}}</math>
-which gives us the value of <math>a \cdot b</math>, which is  <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
+~ lovelearning999
-==Solution 4==
+==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>.
-First observe that the given squares have areas <math>4</math> and <math>5</math>.
+We can then create the equation <math>a+b=\sqrt5</math>.
-Then observe that the 4 triangles with side lengths <math>a</math> and <math>b</math> have a combined area of <math>5-4=1</math>.
+Using the same tactic we get that the side length of the inner square is <math>2</math>. By the Pythagorean Theorem,
-We have, that <math>4\cdot\frac{ab}{2}=2ab</math> is the total area of the 4 triangles in terms of <math>a</math> and <math>b</math>.
+<math>a^2 + b^2 = 2^2</math>.
-Since <math>2ab=1</math>, we divide by two getting <math>a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}</math>
+We can then express this expression as
-LOOK AT THIS OK
+<math>(a+b)^2 - 2ab = 4</math>.
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
+We recall that <math>a+b=\sqrt5</math> and substitute it into our current equation:
+<math>(\sqrt5)^2 - 2ab = 4</math>.
-copy and paste
+We further simplify this to <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
+==Solution 3 (Pythagorean Theorem & Vieta's Formula)==
+Proceed as in solution 3 until you get the equation <math>2a^2-2a\sqrt5+1=0</math>. The two solutions to this equation will be the long and short side of the triangle we are looking for, or <math>a</math> and <math>b</math> from the problem statement. Using Vieta's formula, their product is the constant term divided by the quadratic term, or <math>\boxed{\textbf{(C)} \frac {1}{2}}</math>.
+~nbeaudrot
+==Video Solution by OmegaLearn==
+https://youtu.be/j3QSD5eDpzU?t=2
+==Video Solution 2==
+https://youtu.be/MhxGq1sSA6U ~savannahsolver
+~ pi_is_3.14
 ==See Also==
 {{AMC8 box|year=2012|num-b=24|after=Last Problem}}
 {{MAA Notice}}

2012 AMC 8 (Problems • Answer Key • Resources)
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