Difference between revisions of "2018 AMC 10B Problems/Problem 17"
(→Solution 4) |
|||
Line 4: | Line 4: | ||
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | \begin{figure}[!ht] | ||
+ | \centering | ||
+ | \resizebox{1\textwidth}{!}{% | ||
+ | \begin{circuitikz} | ||
+ | \tikzstyle{every node}=[font=\LARGE] | ||
+ | \draw [](3.75,15.5) to[short] (13.75,15.5); | ||
+ | \draw [](13.75,15.5) to[short] (13.75,10.75); | ||
+ | \draw [](3.75,15.5) to[short] (3.75,10.75); | ||
+ | \draw [](3.75,10.75) to[short] (13.75,10.75); | ||
+ | \node [font=\LARGE] at (3.5,15.75) {P}; | ||
+ | \node [font=\LARGE] at (14,15.75) {Q}; | ||
+ | \node [font=\LARGE] at (14.25,10.25) {R}; | ||
+ | \node [font=\LARGE] at (3.5,10.25) {S}; | ||
+ | \node [font=\LARGE] at (14.25,13) {6}; | ||
+ | \node [font=\LARGE] at (8.75,15.75) {8}; | ||
+ | \node [font=\LARGE] at (14.5,14) {C}; | ||
+ | \node [font=\LARGE] at (14.25,11.5) {D}; | ||
+ | \node [font=\LARGE] at (11.5,9.75) {E}; | ||
+ | \node [font=\LARGE] at (13.75,11.75) {.}; | ||
+ | \node [font=\LARGE] at (6.25,10.75) {.}; | ||
+ | \node [font=\LARGE] at (13.75,14.25) {.}; | ||
+ | \node [font=\LARGE] at (7,15.5) {.}; | ||
+ | \node [font=\LARGE] at (11.25,15.5) {.}; | ||
+ | \node [font=\LARGE] at (7,15.75) {A}; | ||
+ | \node [font=\LARGE] at (11.25,15.75) {B}; | ||
+ | \node [font=\LARGE] at (6.25,10) {F}; | ||
+ | \node [font=\LARGE] at (3.75,11.75) {.}; | ||
+ | \node [font=\LARGE] at (3.75,14.5) {.}; | ||
+ | \node [font=\LARGE] at (3.25,14.25) {G}; | ||
+ | \node [font=\LARGE] at (3.25,11.5) {H}; | ||
+ | \draw [, line width=0.8pt](7,15.5) to[short] (11.25,15.5); | ||
+ | \draw [line width=0.8pt, short] (11.25,15.5) .. controls (12.5,15) and (12.5,15) .. (13.75,14.25); | ||
+ | \draw [line width=0.8pt, short] (13.75,14.5) .. controls (13.75,13.25) and (13.75,13.25) .. (13.75,11.75); | ||
+ | \draw [line width=0.8pt, short] (13.75,11.75) .. controls (12.75,11.25) and (12.75,11.25) .. (11.5,10.75); | ||
+ | \draw [line width=0.8pt, short] (11.5,10.75) .. controls (9,10.75) and (9,10.75) .. (6.25,10.75); | ||
+ | \draw [line width=0.8pt, short] (6.25,10.75) .. controls (5,11.25) and (5,11.25) .. (3.75,11.75); | ||
+ | \draw [line width=0.8pt, short] (3.75,12) .. controls (3.75,13.25) and (3.75,13.25) .. (3.75,14.5); | ||
+ | \draw [line width=0.8pt, short] (3.75,14.5) .. controls (5.5,15) and (5.5,15) .. (7,15.5); | ||
+ | \draw [line width=0.8pt, short] (5,15) .. controls (5.25,15) and (5.25,15) .. (5.25,14.75); | ||
+ | \draw [line width=0.8pt, short] (8.75,15.75) .. controls (8.75,15.5) and (8.75,15.5) .. (8.75,15.25); | ||
+ | \draw [line width=0.8pt, short] (12.75,15) .. controls (12.75,15) and (12.75,15) .. (12.5,14.75); | ||
+ | \draw [line width=0.8pt, short] (13.5,13) .. controls (13.75,13) and (13.75,13) .. (14,13); | ||
+ | \draw [line width=0.8pt, short] (12.5,11.5) .. controls (12.75,11.25) and (12.75,11.25) .. (13,11); | ||
+ | \draw [line width=0.8pt, short] (9,11) .. controls (9,10.75) and (9,10.75) .. (9,10.5); | ||
+ | \draw [line width=0.8pt, short] (5.25,11.5) .. controls (5,11.25) and (5,11.25) .. (4.75,11); | ||
+ | \draw [line width=0.8pt, short] (5,15) .. controls (5.25,14.75) and (5.25,14.75) .. (5.5,14.5); | ||
+ | \draw [line width=0.8pt, short] (12.75,15) .. controls (12.5,14.75) and (12.5,14.75) .. (12.25,14.5); | ||
+ | \end{circuitikz} | ||
+ | }% | ||
+ | |||
+ | \label{fig:my_label} | ||
+ | \end{figure} | ||
+ | |||
+ | ~MC_ADe | ||
== Solution 1== | == Solution 1== |
Revision as of 17:24, 11 November 2023
Contents
[hide]Problem
In rectangle , and . Points and lie on , points and lie on , points and lie on , and points and lie on so that and the convex octagon is equilateral. The length of a side of this octagon can be expressed in the form , where , , and are integers and is not divisible by the square of any prime. What is ?
Diagram
\begin{figure}[!ht]
\centering
\resizebox{1\textwidth}{!}{%
\label{fig:my_label} \end{figure}
~MC_ADe
Solution 1
Let . Then .
Now notice that since we have .
Thus by the Pythagorean Theorem we have which becomes .
Our answer is . (Mudkipswims42)
Solution 2
Denote the length of the equilateral octagon as . The length of can be expressed as . By the Pythagorean Theorem, we find that: Since , we can say that . We can discard the negative solution, so ~ blitzkrieg21
Solution 3
Let the octagon's side length be . Then and . By the Pythagorean theorem, , so . By expanding the left side and combining the like terms, we get . Solving this using the quadratic formula, , we use , , and , to get one positive solution, , so
Solution 4
Let , or the side of the octagon, be . Then, and . By the Pythagorean Theorem, , or . Multiplying this out, we have . Simplifying, . Dividing both sides by gives . Therefore, using the quadratic formula, we have . Since lengths are always positive, then
~MrThinker
Video Solution
~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.