Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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− | ==Problem | + | ==Problem== |
− | In the cube <math>ABCDEFGH</math> with opposite vertices <math>C</math> and <math>E,</math> <math>J</math> and <math>I</math> are the midpoints of | + | In the cube <math>ABCDEFGH</math> with opposite vertices <math>C</math> and <math>E,</math> <math>J</math> and <math>I</math> are the midpoints of segments <math>\overline{FB}</math> and <math>\overline{HD},</math> respectively. Let <math>R</math> be the ratio of the area of the cross-section <math>EJCI</math> to the area of one of the faces of the cube. What is <math>R^2?</math> |
<asy> | <asy> | ||
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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. | + | 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= \sqrt | + | 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== | ||
+ | |||
+ | This time, instead of using a variable like we did in Solution 1, <math>s</math>, for the side length as in the above solution, choose an easy value for <math>s</math> such as <math>2</math>. In the above solution, <math>s^2</math> cancels out in the end, so ultimately the answers are equivalent. | ||
+ | ~Technodoggo | ||
+ | |||
+ | ==Solution 3 (Coordinate Geometry)== | ||
+ | If the edges of the cube have same lengths <math> a </math>, <math> A </math> is the origin, <math> \vec{AD }</math> is the positive <math> x </math> direction, <math> \vec{AE} </math> is the positive <math> y </math> direction, and <math> \vec{AB }</math> is the positive <math> z </math> direction. Therefore, we have <math> E(0,a,0), I(a,\frac{a}{2},0), C(a,0,a), </math> and <math> J(0,\frac{a}{2},a) </math>. Hence, we can figure out that: | ||
+ | |||
+ | <cmath> |EC|=\sqrt{a^2 + a^2 + a^2}=\sqrt{3}a </cmath> | ||
+ | |||
+ | <cmath> |IJ|=\sqrt{a^2 + 0 + a^2}=\sqrt{2}a </cmath> | ||
+ | |||
+ | Note that <math>EJCI</math> is a rhombus, so <math> R=\frac{\sqrt{6}a^2}{2a^2}=\frac{\sqrt{6}}{2} </math>. Finally, we can see that the answer is <math> R^2=\frac{6}{4}=\boxed{\textbf{(C) } \frac{3}{2}} </math> | ||
+ | |||
+ | ~[[User:Bloggish|Bloggish]] | ||
+ | |||
+ | ==Solution 4 (AMC 10+ tactics)== | ||
+ | |||
+ | 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 | ||
+ | |||
+ | == Solution 5 (Not for use in a time-limited contest) == | ||
+ | |||
+ | We will use the following | ||
+ | |||
+ | Theorem: Suppose we have a quadrilateral with edges of length <math>a,b,c,d</math> (in that order) and diagonals of length <math>p, q</math>. Bretschneider's formula states that the area <math>[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>. | ||
+ | |||
+ | Proof: Here is one of my favorite proofs of this with vector geometry. | ||
+ | |||
+ | Suppose a quadrilateral has sides <math>\vec{a}, \vec{b}, \vec{c}, \vec{d}</math> such that <math>\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}</math> and that the diagonals of the quadrilateral are <math>\vec{p} = \vec{b} + \vec{c} = -\vec{a} - \vec{d}</math> and <math>\vec{q} = \vec{a} + \vec{b} = -\vec{c} - \vec{d}</math>. The area of any such quadrilateral is <math>\frac{1}{2} |\vec{p} \times \vec{q}|</math>. | ||
+ | |||
+ | |||
+ | <math>K = \frac{1}{2} |\vec{p} \times \vec{q}|</math> | ||
+ | |||
+ | Lagrange's Identity states that <math>|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2=|\vec{a}\times\vec{b}|^2 \implies \sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2}=|\vec{a}\times\vec{b}|</math>. Therefore: | ||
+ | |||
+ | <math>K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - (2 \vec{p} \cdot \vec{q})^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 (\vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b})]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{b} \cdot (\vec{a} + \vec{b}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot (\vec{c} + \vec{d}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d} + 2 \vec{a} \cdot \vec{c} + 2 \vec{b} \cdot \vec{c}]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{a} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d}]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - ([(\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - \vec{a}\cdot\vec{a} - \vec{c}\cdot\vec{c}] - [(\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d}) - \vec{b}\cdot\vec{b} - \vec{d}\cdot\vec{d}])^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + (\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - (\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d})]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |\vec{b} + \vec{d}|^2]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |-(\vec{a} + \vec{c})|^2]^2}</math> | ||
+ | |||
+ | <math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2]^2}</math> | ||
+ | |||
+ | Then, if <math>a, b, c, d</math> represent <math>|\vec{a}|, |\vec{b}|, |\vec{c}|, |\vec{d}|</math> (and are thus the side lengths) while <math>p, q</math> represent <math>|\vec{p}|, |\vec{q}|</math> (and are thus the diagonal lengths), the area of a quadrilateral is: | ||
+ | |||
+ | <cmath>K = \frac{1}{4} \sqrt{4p^2q^2 - (b^2 + d^2 - a^2 - c^2)^2}</cmath> <math>\square</math> | ||
+ | |||
+ | Back to the problem. We use vectors. WLOG suppose the cube has sides of length <math>2</math>. Let <math>C</math> be the origin and let <math>\vec{CD}</math> be the <math>x</math> direction, <math>\vec{CG}</math> be the <math>y</math> direction, and <math>\vec{CB}</math> be the <math>z</math> direction. | ||
+ | |||
+ | Then, <math>\vec{CI}=\vec{JE}=\begin{pmatrix} 2\\1\\0\end{pmatrix}</math> and <math>\vec{CJ}=\vec{IE}=\begin{pmatrix} 0\\1\\2\end{pmatrix}</math>. Let <math>\angle JCI=\theta</math>. Dot product gives <math>\cos \theta=\left(\frac{\begin{pmatrix} 2\\1\\0\end{pmatrix} \cdot \begin{pmatrix} 0\\1\\2\end{pmatrix}}{|\begin{pmatrix} 2\\1\\0\end{pmatrix}||\begin{pmatrix} 0\\1\\2\end{pmatrix}|}\right)=\frac 15.</math> | ||
+ | |||
+ | Hence, <math>\vec {IJ}=\sqrt{|\vec{CI}|+|\vec{CJ}|-2\cdot \vec{CI}\cdot \vec{CJ}\cos \theta}=\sqrt{|\begin{pmatrix} 2\\1\\0\end{pmatrix}|+|\begin{pmatrix} 0\\1\\2\end{pmatrix}|-2\cdot \begin{pmatrix} 2\\1\\0\end{pmatrix} \cdot\begin{pmatrix} 0\\1\\2\end{pmatrix} \cdot \frac 15} \iff |\vec{IJ}|=2\sqrt2</math>. | ||
+ | |||
+ | Now, notice that <math>\vec{CE}=\vec{CI}+\vec{CJ}=\begin{pmatrix} 2\\1\\0\end{pmatrix}+\begin{pmatrix} 0\\1\\2\end{pmatrix}=\begin{pmatrix} 2\\2\\2\end{pmatrix}</math> so <math>|\vec{CE}|=|\begin{pmatrix} 2\\2\\2\end{pmatrix}|=2\sqrt3</math>. | ||
+ | |||
+ | Using Bretschneider's formula, we obtain <math>[EJCI]=\frac{1}{4} \cdot \sqrt{4|\vec{IJ}|^2|\vec{CE}|^2-(|\vec{CJ}|^2+|\vec{IE}|^2-|\vec{JE}|^2-|\vec{CI}|^2)^2}=\frac 14\sqrt{4\cdot (2\sqrt2)^2\cdot (2\sqrt3)^2-(5^2+5^2-5^2-5^2)^2}=2\sqrt6</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Hence, the answer is <math>\left(\frac{2\sqrt6}{4}\right)^2=\frac 32</math> so we circle answer choice <math>C</math>. | ||
+ | |||
+ | ~franzliszt (again!) | ||
+ | |||
+ | ==Note== | ||
+ | [[2008 AMC 10A Problems/Problem 21|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 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/FDgcLW4frg8?t=2823 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/IrxtHlbERe8 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 13:25, 14 January 2024
Contents
Problem
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
This time, instead of using a variable like we did in Solution 1, , for the side length as in the above solution, choose an easy value for
such as
. In the above solution,
cancels out in the end, so ultimately the answers are equivalent.
~Technodoggo
Solution 3 (Coordinate Geometry)
If the edges of the cube have same lengths ,
is the origin,
is the positive
direction,
is the positive
direction, and
is the positive
direction. Therefore, we have
and
. Hence, we can figure out that:
Note that is a rhombus, so
. Finally, we can see that the answer is
Solution 4 (AMC 10+ tactics)
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
Solution 5 (Not for use in a time-limited contest)
We will use the following
Theorem: Suppose we have a quadrilateral with edges of length (in that order) and diagonals of length
. Bretschneider's formula states that the area
.
Proof: Here is one of my favorite proofs of this with vector geometry.
Suppose a quadrilateral has sides such that
and that the diagonals of the quadrilateral are
and
. The area of any such quadrilateral is
.
Lagrange's Identity states that . Therefore:
Then, if represent
(and are thus the side lengths) while
represent
(and are thus the diagonal lengths), the area of a quadrilateral is:
Back to the problem. We use vectors. WLOG suppose the cube has sides of length . Let
be the origin and let
be the
direction,
be the
direction, and
be the
direction.
Then, and
. Let
. Dot product gives
Hence, .
Now, notice that so
.
Using Bretschneider's formula, we obtain .
Using Bretschneider's formula again, we can find that .
Hence, the answer is so we circle answer choice
.
~franzliszt (again!)
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
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=2823
~ pi_is_3.14
Video Solution
~savannahsolver
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.