Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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<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== | + | ==Solution 1== |
Note that <math>EJCI</math> is a rhombus by symmetry. | Note that <math>EJCI</math> is a rhombus by symmetry. | ||
Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= s\sqrt 3</math> and <math>JI= s\sqrt 2</math>. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is <math>\frac{s^2\sqrt 6}{2}</math>. This gives <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= s\sqrt 3</math> and <math>JI= s\sqrt 2</math>. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is <math>\frac{s^2\sqrt 6}{2}</math>. This gives <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We can solve this with 3D Cartesian coordinates. Assume WLOG that the sides of the square are of length <math>2</math>. Let <math>C</math> be the origin and let <math>\vec{CD}</math> be the positive <math>x</math> direction, <math>\vec{CG}</math> be the positive <math>y</math> direction, and <math>\vec{CB}</math> be the positive <math>z</math> direction. We find that <math>I=(2,1,0),J=(0,1,2),E=(2,2,2)</math>. | ||
+ | |||
+ | Notice that <math>CI=CJ=EI=EJ=\sqrt{5}</math> so <math>EJCI</math> is a rhombus. Furthermore, by the distance formula, <math>IJ=\sqrt8</math>. | ||
+ | |||
+ | By the Law of Cosines on <math>\triangle JCI</math> we have <math>\cos \angle JCI=\frac{\sqrt5^2+ \sqrt5^2-\sqrt8^2}{2\cdot \sqrt5\cdot \sqrt5}=\frac 15</math>. By the Law of Cosines on <math>\triangle JEI</math> we have <math>\cos \angle JEI=\frac{\sqrt5^2+ \sqrt5^2-\sqrt8^2}{2\cdot \sqrt5\cdot \sqrt5}=\frac 15</math>. | ||
+ | |||
+ | Bretschneider's formula states given a quadrilateral <math>ABCD</math> with sides <math>a,b,c,d</math> then <cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}</cmath>where <math>s=\frac{a+b+c+d}2</math>. Using this formula, we find that<cmath>[EJCI]=\sqrt{(2\sqrt5-\sqrt5)(2\sqrt5-\sqrt5)(2\sqrt5-\sqrt5)(2\sqrt5-\sqrt5)-\sqrt 5^4(\tfrac 15)^2}=2\sqrt{6}.</cmath> | ||
+ | |||
+ | Using Bretschneider's formula again, we can find that <math>[ABCD]=\sqrt{(4-2)(4-2)(4-2)(4-2)-2^4\cdot \cos \left(\frac{90^\circ+90^\circ}2\right)}=\sqrt{16}=4</math>. | ||
+ | |||
+ | The answer is thus <math>\left(\frac{2\sqrt6}{4}\right)^2=\frac 32</math> so we circle answer choice <math>C</math>. | ||
+ | |||
+ | ~franzliszt | ||
==Note== | ==Note== |
Revision as of 18:36, 6 November 2020
Contents
[hide]Problem 24
In the cube with opposite vertices
and
and
are the midpoints of segments
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 1
Note that is a rhombus by symmetry.
Let the side length of the cube be
. By the Pythagorean theorem,
and
. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is
. This gives
. Thus
Solution 2
We can solve this with 3D Cartesian coordinates. Assume WLOG that the sides of the square are of length . Let
be the origin and let
be the positive
direction,
be the positive
direction, and
be the positive
direction. We find that
.
Notice that so
is a rhombus. Furthermore, by the distance formula,
.
By the Law of Cosines on we have
. By the Law of Cosines on
we have
.
Bretschneider's formula states given a quadrilateral with sides
then
where
. Using this formula, we find that
Using Bretschneider's formula again, we can find that .
The answer is thus so we circle answer choice
.
~franzliszt
Note
Problem 21 of the 2008 AMC 10A was nearly identical to this question, except that in this question you have to look for the square of the area, not the actual area.
Video Solution
https://www.youtube.com/watch?v=04pV_rZw8bg - Happytwin
Video Solution
https://www.youtube.com/watch?v=ji9_6XNxyIc ~ MathEx
See Also
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.