Difference between revisions of "2021 AMC 12B Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>MD= | + | Let <math>MA=a</math> and <math>MD=d.</math> It follows that <math>MC=a+2</math> and <math>MB=a+4.</math> |
+ | |||
+ | As shown below, note that <math>\triangle MAD</math> and <math>\triangle MBC</math> are both right triangles. | ||
+ | <asy> | ||
+ | size(300); | ||
+ | import graph3; | ||
+ | import solids; | ||
+ | |||
+ | currentprojection=orthographic((0.5,-0.25,-0.5)); | ||
+ | triple A, B, C, D, M; | ||
+ | A = (-2sqrt(10),0,0); | ||
+ | B = (-2sqrt(10),-6sqrt(2),0); | ||
+ | C = (0,-6sqrt(2),0); | ||
+ | D = (0,0,0); | ||
+ | M = (0,0,3); | ||
+ | draw(surface(M--A--D--cycle),yellow); | ||
+ | draw(surface(M--B--C--cycle),yellow); | ||
+ | draw(D--A--B--C^^A--M^^B--M^^C--M^^D--M); | ||
+ | draw(C--D,dashed); | ||
+ | dot(A^^B^^C^^D^^M,linewidth(4.5)); | ||
+ | label("$A$",A,2*dir(A-B)); | ||
+ | label("$B$",B,2*dir(B-A)); | ||
+ | label("$C$",C,2*dir(C-D)); | ||
+ | label("$D$",D,2*dir(D-C)); | ||
+ | label("$M$",M,2*dir((1,1,0))); | ||
+ | label("$a$",midpoint(M--A),2*dir((-1,-1,0)),red); | ||
+ | label("$d$",midpoint(M--D),2*dir((1,1,0)),red); | ||
+ | label("$a+2$",midpoint(M--C),2*dir((1,-1,0)),red); | ||
+ | label("$a+4$",midpoint(M--B),2*dir((-1,1,0)),red); | ||
+ | </asy> | ||
+ | By the Pythagorean Theorem, we have | ||
+ | <cmath>\begin{alignat*}{6} | ||
+ | AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \ | ||
+ | BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. | ||
+ | \end{alignat*}</cmath> | ||
+ | Since <math>AD=BC</math> in rectangle <math>ABCD,</math> we equate the expressions for <math>AD^2</math> and <math>BC^2,</math> then rearrange and factor: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | a^2 | + | a^2 - d^2 &= (a+4)^2 - (a+2)^2 \ |
− | + | a^2 - d^2 &= 4a + 12 \ | |
− | a^2- | + | a^2 - 4a - d^2 &= 12 \ |
− | (a-2 | + | (a-2)^2 - d^2 &= 16 \ |
+ | (a+d-2)(a-d-2) &= 16. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | from which <math>(a, | + | As <math>a+d-2</math> and <math>a-d-2</math> have the same parity, we get <math>a+d-2=8</math> and <math>a-d-2=2,</math> from which <math>(a,d)=(7,3).</math> |
+ | |||
+ | Applying the Pythagorean Theorem to right <math>\triangle MAD</math> and right <math>\triangle MCD,</math> we obtain <math>AD=2\sqrt{10}</math> and <math>CD=6\sqrt2,</math> respectively. | ||
− | ~Lopkiloinm | + | Let the brackets denote areas. Together, the volume of pyramid <math>MABCD</math> is <cmath>\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.</cmath> |
+ | ~Lopkiloinm ~MRENTHUSIASM | ||
==Solution 2== | ==Solution 2== | ||
− | Let <math> | + | Let <math>AD=b</math>, <math>CD=a</math>, <math>MD=x</math>, <math>MC=t</math>. It follows that <math>MA=t-2</math> and <math>MB=t+2</math>. |
− | <math> | ||
We have three equations: | We have three equations: | ||
Line 26: | Line 64: | ||
b^2 + x^2 &= t^2 - 4t + 4. | b^2 + x^2 &= t^2 - 4t + 4. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Substituting the first and third equations into the second equation, we get: | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | t^2 - 8t - x^2 &= 0 | + | t^2 - 8t - x^2 &= 0 \ |
− | (t-4)^2 - x^2 &= 16 | + | (t-4)^2 - x^2 &= 16 \ |
(t-4-x)(t-4+x) &= 16. | (t-4-x)(t-4+x) &= 16. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
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~jamess2022 (burntTacos) | ~jamess2022 (burntTacos) | ||
+ | |||
+ | |||
+ | ==Video Solution (Fast! Just 4 min!)== | ||
+ | https://youtu.be/Bo2EvRZdRnA | ||
+ | |||
+ | ~<i>Education, the Study of Everything</i> | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== | ||
Line 44: | Line 88: | ||
== Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid) == | == Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid) == | ||
https://youtu.be/4_Oqp_ECLRw | https://youtu.be/4_Oqp_ECLRw | ||
+ | |||
+ | ~pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:38, 23 July 2023
Contents
[hide]Problem
Let be a rectangle and let be a segment perpendicular to the plane of . Suppose that has integer length, and the lengths of and are consecutive odd positive integers (in this order). What is the volume of pyramid
Solution 1
Let and It follows that and
As shown below, note that and are both right triangles. By the Pythagorean Theorem, we have Since in rectangle we equate the expressions for and then rearrange and factor: As and have the same parity, we get and from which
Applying the Pythagorean Theorem to right and right we obtain and respectively.
Let the brackets denote areas. Together, the volume of pyramid is ~Lopkiloinm ~MRENTHUSIASM
Solution 2
Let , , , . It follows that and .
We have three equations: Substituting the first and third equations into the second equation, we get: Therefore, we have and .
Solving for other values, we get , . The volume is then
~jamess2022 (burntTacos)
Video Solution (Fast! Just 4 min!)
~Education, the Study of Everything
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid)
~pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
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