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 "2012 AMC 8 Problems/Problem 25"

(→‎Solution 2)
 
(12 intermediate revisions by 5 users not shown)
Line 13: Line 13:
 
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>.
 
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 3 (similar to solution 1)==
+
==Solution 2 (Algebra)==
 +
We see that we want <math>ab</math>, so instead of solving for <math>a,b</math>, we find a way to get an expression with <math>ab</math>.
  
−
Since we know that each of the <math>4</math> triangles have side lengths <math>a</math> and <math>b</math>, we can create an equation: the area of the inner square plus the sum of the <math>4</math> triangles equals the area of the outer square.
+
By Triple Perpendicularity Model,  
  
−
<cmath> 4 + 2ab = 5</cmath>
+
all four triangles are congruent.
  
−
which gives us the value of <math>a \cdot b</math>, which is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>.
+
By Pythagorean's Theorem,
  
−
==Solution 4==
+
<math>\sqrt{a^2+b^2} = \sqrt{4}</math>
  
−
First, observe that the given squares have areas <math>4</math> and <math>5</math>.
+
Thus, <math>\sqrt{a^2+b^2} = 2</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>.
+
As <math>a+b=\sqrt{5}</math>,
  
−
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+2ab+b^2 = 5</math>
  
−
Since <math>2ab=1</math>, we divide by two getting <math>a \cdot b=\boxed{\textbf{(C) }\frac{1}{2}}</math>
+
So, <math>\sqrt{5-2ab} = 2</math>
  
−
==Video Solution by Punxsutawney Phil==
+
Simplifying,
−
https://youtu.be/RyKWp2YDHJM
 
  
−
~sugar_rush
+
<math>5-2ab = 4</math>
  
−
https://www.youtube.com/watch?v=QEwZ_17PQ6Q  ~David
+
<math>-2ab=-1</math>
  
 +
<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==
 
==Video Solution 2==
 
https://youtu.be/MhxGq1sSA6U ~savannahsolver
 
https://youtu.be/MhxGq1sSA6U ~savannahsolver
Line 47: Line 90:
  
 
==See Also==
 
==See Also==
−
{{AMC8 box|year=2012|num-b=24|after=Last Problem}}
+
{{AMC8 box|year=2012|num-b=24|after=None}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
None
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_25&oldid=237781"