Difference between revisions of "2018 AMC 8 Problems/Problem 24"
(→Problem 24) |
|||
Line 32: | Line 32: | ||
<math>\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}</math> | <math>\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Note that <math>EJCI</math> is a rhombus. | ||
+ | Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | ||
+ | |||
{{AMC8 box|year=2018|num-b=23|num-a=25}} | {{AMC8 box|year=2018|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:22, 21 November 2018
Problem 24
In the cube with opposite vertices
and
and
are the midpoints of edges
and
respectively. Let
be the ratio of the area of the cross-section
to the area of one of the faces of the cube. What is
Solution
Note that is a rhombus.
Let the side length of the cube be
. By the Pythagorean theorem,
and
. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is
.
. Thus
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.