Difference between revisions of "2018 AMC 10B Problems/Problem 17"
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== 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>\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> | 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> | ||
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+ | ==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> | ||
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+ | ~MrThinker | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 18:54, 11 November 2023
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 |
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