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 "2018 AMC 10B Problems/Problem 17"

(→‎Solution 2)
(→‎Solution 1)
 
(16 intermediate revisions by 9 users not shown)
Line 11: Line 11:
 
Now notice that since <math>CD=8-2x</math> we have <math>QC=DR=x-1</math>.
 
Now notice that since <math>CD=8-2x</math> we have <math>QC=DR=x-1</math>.
  
−
Thus by the Pythagorean Theorem we have <math>x^2+(x-1)^2=(8-2x)^2</math> which becomes <math>2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}</math>.
+
Thus by the Pythagorean Theorem we have <math>x^2+(x-1)^2=(8-2x)^2</math> which becomes <math>2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}</math>
  
 
Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42)
 
Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42)
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Denote the length of the equilateral octagon as <math>x</math>. The length of <math>\overline{BQ}</math> can be expressed as <math>\frac{8-x}{2}</math>. By Pythagoras, we find that:
+
Denote the length of the equilateral octagon as <math>x</math>. The length of <math>\overline{BQ}</math> can be expressed as <math>\frac{8-x}{2}</math>. By the Pythagorean Theorem, we find that:
−
<cmath>\left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\implies \overline{CQ}=\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}</cmath>  
+
<cmath>\left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\implies \overline{CQ}=\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}</cmath>  
 
Since <math>\overline{CQ}=\overline{DR}</math>, we can say that <math>x+2\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}=6\implies x=-7\pm3\sqrt{11}</math>. We can discard the negative solution, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math> ~ blitzkrieg21
 
Since <math>\overline{CQ}=\overline{DR}</math>, we can say that <math>x+2\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}=6\implies x=-7\pm3\sqrt{11}</math>. We can discard the negative solution, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math> ~ blitzkrieg21
 +
 +
== Solution 3 ==
 +
Let the octagon's side length be <math>x</math>. Then <math>PH = \frac{6 - x}{2}</math> and <math>PA = \frac{8 - x}{2}</math>. By the Pythagorean theorem, <math>PH^2 + PA^2 = HA^2</math>, so <math>\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2</math>. By expanding the left side and  combining the like terms, we get <math>\frac{x^2}{2} - 7x + 25 = x^2 \implies -\frac{x^2}{2} - 7x + 25 = 0</math>. Solving this using the quadratic formula, <math>\frac{-b \pm \sqrt{b^2-4ac}}{2a}</math>, we use <math>a = -\frac{1}{2}</math>, <math>b = -7</math>, and <math>c = 25</math>, to get one positive solution, <math>x=-7+3\sqrt{11}</math>, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math>
 +
 +
==Solution 4==
 +
Let <math>AB</math>, or the side of the octagon, be <math>x</math>. Then, <math>BQ = \left(\frac{8-x}{2}\right)</math> and <math>CQ = \left(\frac{6-x}{2}\right)</math>. By the [[Pythagorean Theorem]], <math>BQ^2+CQ^2=x^2</math>, or <math>\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2</math>. Multiplying this out, we have <math>x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}</math>. Simplifying, <math>-2x^2-28x+100=0</math>. Dividing both sides by <math>-2</math> gives <math>x^2+14x-50=0</math>. Therefore, using the [[quadratic formula]], we have <math>x=-7 \pm 3\sqrt{11}</math>. Since lengths are always positive, then <math>x=-7+3\sqrt{11} \Rightarrow k+m+n=-7+3+11=\boxed{\textbf{(B)}\ 7}</math>
 +
 +
~MrThinker
 +
 +
==Video Solution==
 +
https://youtu.be/8sts_hn7cpQ
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==

Latest revision as of 19:37, 2 September 2024

Problem

In rectangle $PQRS$, $PQ=8$ and $QR=6$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, points $E$ and $F$ lie on $\overline{RS}$, and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$, where $k$, $m$, and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$?

$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$

Solution 1

Let $AP=BQ=x$. Then $AB=8-2x$.

Now notice that since $CD=8-2x$ we have $QC=DR=x-1$.

Thus by the Pythagorean Theorem we have $x^2+(x-1)^2=(8-2x)^2$ which becomes $2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}$

Our answer is $8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}$. (Mudkipswims42)

Solution 2

Denote the length of the equilateral octagon as $x$. The length of $\overline{BQ}$ can be expressed as $\frac{8-x}{2}$. By the Pythagorean Theorem, we find that: \[\left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\implies \overline{CQ}=\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}\] Since $\overline{CQ}=\overline{DR}$, we can say that $x+2\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}=6\implies x=-7\pm3\sqrt{11}$. We can discard the negative solution, so $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$ ~ blitzkrieg21

Solution 3

Let the octagon's side length be $x$. Then $PH = \frac{6 - x}{2}$ and $PA = \frac{8 - x}{2}$. By the Pythagorean theorem, $PH^2 + PA^2 = HA^2$, so $\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2$. By expanding the left side and combining the like terms, we get $\frac{x^2}{2} - 7x + 25 = x^2 \implies -\frac{x^2}{2} - 7x + 25 = 0$. Solving this using the quadratic formula, $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we use $a = -\frac{1}{2}$, $b = -7$, and $c = 25$, to get one positive solution, $x=-7+3\sqrt{11}$, so $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$

Solution 4

Let $AB$, or the side of the octagon, be $x$. Then, $BQ = \left(\frac{8-x}{2}\right)$ and $CQ = \left(\frac{6-x}{2}\right)$. By the Pythagorean Theorem, $BQ^2+CQ^2=x^2$, or $\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2$. Multiplying this out, we have $x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}$. Simplifying, $-2x^2-28x+100=0$. Dividing both sides by $-2$ gives $x^2+14x-50=0$. Therefore, using the quadratic formula, we have $x=-7 \pm 3\sqrt{11}$. Since lengths are always positive, then $x=-7+3\sqrt{11} \Rightarrow k+m+n=-7+3+11=\boxed{\textbf{(B)}\ 7}$

~MrThinker

Video Solution

https://youtu.be/8sts_hn7cpQ

~IceMatrix

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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=2018_AMC_10B_Problems/Problem_17&oldid=226260"