2018 AMC 10B Problems/Problem 17
Contents
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
import olympiad; \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
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 |
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