Difference between revisions of "2018 AMC 10B Problems/Problem 17"
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<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> | ||
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== Solution 1== | == Solution 1== | ||
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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) |
Latest revision as of 19:37, 2 September 2024
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
?
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.