Difference between revisions of "2021 AMC 12B Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (Undo revision 180241 by Puffer13 (talk) I did make the diagram and revised the solution to be clearer, so we coauthored.) (Tag: Undo) |
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~jamess2022 (burntTacos) | ~jamess2022 (burntTacos) | ||
+ | |||
+ | ==Solution 3== | ||
+ | First and foremost, the diagram must be drawn. | ||
+ | <asy> | ||
+ | import graph; | ||
+ | |||
+ | size(250); | ||
+ | |||
+ | pair A = (0,0); | ||
+ | pair B = (4.5,0); | ||
+ | pair C = (6,1.1); | ||
+ | pair D = (1.5,1.1); | ||
+ | pair M = (1.5,3.45); | ||
+ | |||
+ | draw(A--B); | ||
+ | draw(C--B); | ||
+ | draw(A--D, dashed); | ||
+ | draw(C--D, dashed); | ||
+ | draw(M--A); | ||
+ | draw(M--B); | ||
+ | draw(M--C); | ||
+ | draw(M--D, dashed); | ||
+ | |||
+ | draw((1.5,1.3)--(1.7,1.3)--(1.7,1.1)); | ||
+ | |||
+ | dot(A,linewidth(4.5)); | ||
+ | dot(B,linewidth(4.5)); | ||
+ | dot(C,linewidth(4.5)); | ||
+ | dot(D,linewidth(4.5)); | ||
+ | dot(M,linewidth(4.5)); | ||
+ | |||
+ | label("$A$", A, W); | ||
+ | label("$B$", B, SE); | ||
+ | label("$C$", C, E); | ||
+ | label("$D$", D, NW); | ||
+ | label("$M$", M, N); | ||
+ | |||
+ | label("$x$", midpoint(M--A), NW, red); | ||
+ | label("$x+4$", midpoint(M--B), SE, red); | ||
+ | label("$x+2$", midpoint(M--C), NE, red); | ||
+ | label("$a$", midpoint(A--D), W, red); | ||
+ | label("$b$", midpoint(A--B), S, red); | ||
+ | label("$I$", midpoint(M--D), E, red); | ||
+ | </asy> | ||
+ | With the diagram, an impulse to utilize Pythagorean theorem is created. Thereby, the shaded right triangles may be used. | ||
+ | <asy> | ||
+ | import graph; | ||
+ | |||
+ | size(250); | ||
+ | |||
+ | pair A = (0,0); | ||
+ | pair B = (4.5,0); | ||
+ | pair C = (6,1.1); | ||
+ | pair D = (1.5,1.1); | ||
+ | pair M = (1.5,3.45); | ||
+ | |||
+ | draw(A--B); | ||
+ | draw(C--B); | ||
+ | draw(A--D, dashed); | ||
+ | draw(C--D, dashed); | ||
+ | draw(D--B, dashed); | ||
+ | draw(M--A); | ||
+ | draw(M--B); | ||
+ | draw(M--C); | ||
+ | draw(M--D, dashed); | ||
+ | |||
+ | draw(M--A--D--cycle, blue); | ||
+ | draw(M--B--D--cycle, green); | ||
+ | draw(M--C--D--cycle, red); | ||
+ | |||
+ | draw((1.5,1.3)--(1.7,1.3)--(1.7,1.1)); | ||
+ | |||
+ | dot(A,linewidth(4.5)); | ||
+ | dot(B,linewidth(4.5)); | ||
+ | dot(C,linewidth(4.5)); | ||
+ | dot(D,linewidth(4.5)); | ||
+ | dot(M,linewidth(4.5)); | ||
+ | |||
+ | label("$A$", A, W); | ||
+ | label("$B$", B, SE); | ||
+ | label("$C$", C, E); | ||
+ | label("$D$", D, NW); | ||
+ | label("$M$", M, N); | ||
+ | |||
+ | label("$x$", midpoint(M--A), NW, red); | ||
+ | label("$x+4$", midpoint(M--B), SE, red); | ||
+ | label("$x+2$", midpoint(M--C), NE, red); | ||
+ | label("$a$", midpoint(A--D), W, red); | ||
+ | label("$b$", midpoint(A--B), S, red); | ||
+ | label("$I$", midpoint(M--D), E, red); | ||
+ | </asy> | ||
+ | |||
+ | The following equations could be obtained from Pythagorean Theorem. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | I^2+a^2&=x^2 \\ | ||
+ | I^2+b^2&=(x+2)^2 \\ | ||
+ | I^2+a^2+b^2&=(x+4)^2 \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Because <math>x</math> and <math>I</math> are integers, writing an equation in terms of only <math>x</math> and <math>I</math> may be beneficial for trial and error. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | x^2+(x+2)^2&=(x+4)^2+I^2 \\ | ||
+ | 2x^2+4x+4&=x^2+8x+16+I^2 \\ | ||
+ | x^2-4x-12&=I^2 \\ | ||
+ | (x-6)(x+2)&=I^2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | <math>x=7</math> seems valid since <math>I=3</math> when <math>x=7</math>. Therefore, when <math>x=7</math>, <math>a=2\sqrt{10}</math> and <math>b=6\sqrt{2}</math>. We could double check that <math>40+72+9=11^2</math>. Thereby, | ||
+ | <cmath> | ||
+ | \frac{a\cdot b\cdot I}{3}=\frac{2\sqrt{10}\cdot6\sqrt{2}\cdot3}{3}=\boxed{\textbf{(A) }24\sqrt5}. | ||
+ | </cmath> | ||
+ | |||
+ | ~MaPhyCom | ||
+ | |||
+ | ==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== |
Latest revision as of 05:13, 4 June 2025
Contents
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)
Solution 3
First and foremost, the diagram must be drawn.
With the diagram, an impulse to utilize Pythagorean theorem is created. Thereby, the shaded right triangles may be used.
The following equations could be obtained from Pythagorean Theorem.
Because
and
are integers, writing an equation in terms of only
and
may be beneficial for trial and error.
seems valid since
when
. Therefore, when
,
and
. We could double check that
. Thereby,
~MaPhyCom
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.