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 "2005 AMC 10A Problems/Problem 20"

m (Adjusted problem statement to match the official version)
(Improved formatting, grammar, and explanations)
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
An equilangular octagon has four sides of length <math>1</math> and four sides of length <math>\sqrt{2}/2</math>, arranged so that no two consecutive sides have the same length. What is the area of the octagon?
+
An equiangular octagon has four sides of length <math>1</math> and four sides of length <math>\sqrt{2}/2</math>, arranged so that no two consecutive sides have the same length. What is the area of the octagon?
  
 
<math>
 
<math>
−
\textbf{(A) } \frac{7}{2}\qquad \textbf{(B) } \frac{7\sqrt{2}}{2}\qquad \textbf{(C) } \frac{5+4\sqrt{2}}{2}\qquad \textbf{(D) } \frac{4+5\sqrt{2}}{2}\qquad \textbf{(E) } 7
+
\textbf{(A) } \frac{7}{2}\qquad \textbf{(B) } \frac{7\sqrt{2}}{2}\qquad \textbf{(C) } \frac{5+4\sqrt{2}}{2}\qquad \textbf{(D) } \frac{4+5\sqrt{2}}{2}\qquad \textbf{(E) } 7
 
</math>
 
</math>
  
 
==Solution 1==
 
==Solution 1==
−
The area of the octagon can be divided up into <math>5</math> squares with side <math>\frac{\sqrt2}2</math> and <math>4</math> right triangles, which are half the area of each of the squares.
+
The sum of the octagon's angles is <math>180\cdot(8-2)^{\circ} = 1080^{\circ}</math>, so since it is equiangular, each angle is <math>\frac{1080^{\circ}}{8} = 135^{\circ}</math>, i.e. the same as in a regular octagon. This means that this octagon, just like with a regular octagon, can be divided into squares (whose diagonals form <math>45^{\circ}</math> angles) and right triangles.
  
−
Therefore, the area of the octagon is equal to the area of <math>5+4\left(\frac12\right)=7</math> squares.
+
In particular, as shown in the diagram below, the area of this octagon can be divided up into <math>5</math> squares of side length <math>\frac{\sqrt{2}}{2}</math> and <math>4</math> right triangles, each of which has half the area of each of the squares (since each square has diagonal <math>\frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1</math>, so each right triangle is congruent by side-side-side to one of the triangles formed by slicing a square along one of its diagonals).
  
−
The area of each square is <math>\left(\frac{\sqrt2}2\right)^2=\frac12</math>, so the area of <math>7</math> squares is <math>\boxed{\textbf{(A) }\frac72}</math>.
+
Therefore, the area of the octagon is equal to the area of <math>5 + 4 \cdot \frac{1}{2} = 7</math> of the squares, and the area of each square is simply <math>\left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}</math>, so the answer is is <math>7 \cdot \frac{1}{2} = \boxed{\textbf{(A) } \frac{7}{2}}</math>.
  
 
<asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H);  draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy>
 
<asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H);  draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy>
  
 
==Solution 2==
 
==Solution 2==
−
Using the diagram from above, we can extend the sides of length <math>1</math> to form four right triangles and the octagon, all inside a square. If you wanted, you could also extend the sides of length <math>\frac{\sqrt{2}}{2}</math> and get the same answer, but that would make things slightly harder. The right triangles are <math>45-45-90</math> triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>, so the side length is <math>\frac{1}{2}</math>. Thus, the area of the larger square is <math>2 \cdot 2 = 4</math>, and the area of the four right triangles combined is <math>\frac{1}{2}</math>, so the area of the octagon is <math>4-\frac{1}{2}=\boxed{\textbf{(A) }\frac{7}{2}}</math>
+
Using the same diagram as in Solution 1, we can extend the sides of length <math>1</math> to form a bounding square around the octagon, with the region inside the square but outside the octagon consisting of <math>4</math> right triangles. (We could also extend the sides of length <math>\frac{\sqrt{2}}{2}</math> to give a similar bounding square, but this would make the computations slightly harder.)
 +
 
 +
As in Solution 1, since all the interior angles of the octagon are <math>135^{\circ}</math>, the right triangles are all <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>. Thus the length of their other legs is <math>\frac{\left(\frac{\sqrt{2}}{2}\right)}{\sqrt{2}} = \frac{1}{2}</math>, so the bounding square has total side length <math>1 + 2 \cdot \frac{1}{2} = 2</math>, and hence area <math>2^2 = 4</math>. Each of the right triangles has area <math>\frac{1}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{8}</math>, so we deduce that the area of the octagon is <math>4 - 4 \cdot \frac{1}{8} = \boxed{\textbf{(A) } \frac{7}{2}}</math>.
  
 
edited by mobius247
 
edited by mobius247
  
 
==Solution 3==
 
==Solution 3==
−
Refer to the following diagram:
+
As shown in the diagram below, we can also divide the octagon into squares and right triangles using horizontal and vertical lines, rather than the lines at a <math>45^{\circ}</math> angle that were used in Solutions 1 and 2.
  
 
[[File:AMC10_2005A_P20.png|500px]]
 
[[File:AMC10_2005A_P20.png|500px]]
  
−
(Picture made on Geogebra)
+
(Diagram made using Geogebra)
−
 
 
−
 
 
−
Note that each square has area <math>\frac14</math>, and each triangle has area <math>\frac18</math>. The total area is <math>12\cdot\frac14+4\cdot\frac18=\frac72 \Longrightarrow \boxed{\textbf{(A) } \frac72}</math>.
 
−
 
 
  
−
Remark: This solution requires careful drawing, and also realising that connecting lines leads to squares and isosceles right triangles (notice the <math>\sqrt{2}</math> and realise that it is the hypotenuse of a <math>45-45-90</math> triangle with side length ratios <math>1:1:\sqrt{2}</math>.).
+
For the same reasons as in Solutions 1 and 2, the octagon's <math>135^{\circ}</math> interior angles mean that the right triangles are <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>, so their other legs have length <math>\frac{\left(\frac{\sqrt{2}}{2}\right)}{\sqrt{2}} = \frac{1}{2}</math>. Accordingly, their areas are each <math>\frac{1}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{8}</math>, while by symmetry, each of the squares also has side length <math>\frac{1}{2}</math>, and thus area <math>\left(\frac{1}{2}\right)^2 = \frac{1}{4}</math>.
  
 +
Hence, as the octagon consists of <math>12</math> squares and <math>4</math> right triangles, its total area is <math>12\cdot\frac{1}{4}+4\cdot\frac{1}{8} = \boxed{\textbf{(A) } \frac{7}{2}}</math>.
  
 
~JH. L
 
~JH. L
  
 
==Video Solution==
 
==Video Solution==
−
CHECK OUT Video Solution: https://youtu.be/rwPFZnYk9V8
+
https://youtu.be/rwPFZnYk9V8
  
 
==See Also==
 
==See Also==

Latest revision as of 02:03, 2 July 2025

Problem

An equiangular octagon has four sides of length $1$ and four sides of length $\sqrt{2}/2$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?

$\textbf{(A) } \frac{7}{2}\qquad \textbf{(B) } \frac{7\sqrt{2}}{2}\qquad \textbf{(C) } \frac{5+4\sqrt{2}}{2}\qquad \textbf{(D) } \frac{4+5\sqrt{2}}{2}\qquad \textbf{(E) } 7$

Solution 1

The sum of the octagon's angles is $180\cdot(8-2)^{\circ} = 1080^{\circ}$, so since it is equiangular, each angle is $\frac{1080^{\circ}}{8} = 135^{\circ}$, i.e. the same as in a regular octagon. This means that this octagon, just like with a regular octagon, can be divided into squares (whose diagonals form $45^{\circ}$ angles) and right triangles.

In particular, as shown in the diagram below, the area of this octagon can be divided up into $5$ squares of side length $\frac{\sqrt{2}}{2}$ and $4$ right triangles, each of which has half the area of each of the squares (since each square has diagonal $\frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1$, so each right triangle is congruent by side-side-side to one of the triangles formed by slicing a square along one of its diagonals).

Therefore, the area of the octagon is equal to the area of $5 + 4 \cdot \frac{1}{2} = 7$ of the squares, and the area of each square is simply $\left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}$, so the answer is is $7 \cdot \frac{1}{2} = \boxed{\textbf{(A) } \frac{7}{2}}$.

[asy] pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); [/asy]

Solution 2

Using the same diagram as in Solution 1, we can extend the sides of length $1$ to form a bounding square around the octagon, with the region inside the square but outside the octagon consisting of $4$ right triangles. (We could also extend the sides of length $\frac{\sqrt{2}}{2}$ to give a similar bounding square, but this would make the computations slightly harder.)

As in Solution 1, since all the interior angles of the octagon are $135^{\circ}$, the right triangles are all $45^{\circ}-45^{\circ}-90^{\circ}$ triangles with hypotenuse $\frac{\sqrt{2}}{2}$. Thus the length of their other legs is $\frac{\left(\frac{\sqrt{2}}{2}\right)}{\sqrt{2}} = \frac{1}{2}$, so the bounding square has total side length $1 + 2 \cdot \frac{1}{2} = 2$, and hence area $2^2 = 4$. Each of the right triangles has area $\frac{1}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{8}$, so we deduce that the area of the octagon is $4 - 4 \cdot \frac{1}{8} = \boxed{\textbf{(A) } \frac{7}{2}}$.

edited by mobius247

Solution 3

As shown in the diagram below, we can also divide the octagon into squares and right triangles using horizontal and vertical lines, rather than the lines at a $45^{\circ}$ angle that were used in Solutions 1 and 2.

AMC10 2005A P20.png

(Diagram made using Geogebra)

For the same reasons as in Solutions 1 and 2, the octagon's $135^{\circ}$ interior angles mean that the right triangles are $45^{\circ}-45^{\circ}-90^{\circ}$ triangles with hypotenuse $\frac{\sqrt{2}}{2}$, so their other legs have length $\frac{\left(\frac{\sqrt{2}}{2}\right)}{\sqrt{2}} = \frac{1}{2}$. Accordingly, their areas are each $\frac{1}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{8}$, while by symmetry, each of the squares also has side length $\frac{1}{2}$, and thus area $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

Hence, as the octagon consists of $12$ squares and $4$ right triangles, its total area is $12\cdot\frac{1}{4}+4\cdot\frac{1}{8} = \boxed{\textbf{(A) } \frac{7}{2}}$.

~JH. L

Video Solution

https://youtu.be/rwPFZnYk9V8

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_20&oldid=253110"