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"

m (→‎Solution 2)
m (→‎Solution 3)
Line 21: Line 21:
  
 
== Solution 3 ==
 
== 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>(\frac{6 - x}{2})^2 + (\frac{8 - x}{2})^2 = x^2</math>. Solving this, we get one positive solution, <math>x=-7+3\sqrt{11}</math>, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math>
+
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>. Solving this, we get one positive solution, <math>x=-7+3\sqrt{11}</math>, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math>
  
 
==See Also==
 
==See Also==

Revision as of 15:40, 17 December 2019

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 $\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2$. Solving this, we get one positive solution, $x=-7+3\sqrt{11}$, so $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$

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=112990"