Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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==Problem== | ==Problem== | ||
+ | |||
+ | The coordinates of <math>\triangle ABC</math> are <math>A(5,7)</math>, <math>B(11,7)</math>, and <math>C(3,y)</math>, with <math>y>7</math>. The area of <math>\triangle ABC</math> is 12. What is the value of <math>y</math>? | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | draw((3,11)--(11,7)--(5,7)--(3,11)); | ||
+ | |||
+ | dot((5,7)); | ||
+ | label("$A(5,7)$",(5,7),S); | ||
+ | |||
+ | dot((11,7)); | ||
+ | label("$B(11,7)$",(11,7),S); | ||
+ | |||
+ | dot((3,11)); | ||
+ | label("$C(3,y)$",(3,11),NW); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | |||
+ | <math>\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
The triangle has base <math>6,</math> which means its height satisfies | The triangle has base <math>6,</math> which means its height satisfies | ||
<cmath>\dfrac{6h}{2}=3h=12.</cmath> | <cmath>\dfrac{6h}{2}=3h=12.</cmath> | ||
This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | draw((5,7)--(11,7)--(3,11)--cycle); | ||
+ | draw((3,11)--(3,7)--(5,7),red); | ||
+ | draw((3,7.5)--(3.5,7.5)--(3.5,7)); | ||
+ | label("$A(5,7)$", (5,7),S); | ||
+ | label("$B(11,7)$", (11,7),S); | ||
+ | label("$C(3,y)$", (3,11),W); | ||
+ | label("$D(3,7)$", (3,7),SW); | ||
+ | </asy> | ||
+ | Label point <math>D(3,7)</math> as the point at which <math>CD\perp DA</math>. We now have <math>[\triangle ABC] = [\triangle BCD] - [\triangle ACD]</math>, where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of <math>\triangle ACD</math> are <math>y-7</math> and <math>5-3=2</math>. The two side lengths of <math>\triangle BCD</math> are <math>y-7</math> and <math>11-3 = 8.</math> Now, | ||
+ | |||
+ | <cmath>[\triangle ABC] = 12 = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 = 3(y-7)</cmath> | ||
+ | |||
+ | Dividing by <math>3</math> gives <math>y -7 = 4,</math> so <math>y = \boxed{\textbf{(D)\ 11}}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Solution 3== | ||
+ | By the Shoelace Theorem, <math>\triangle ABC</math> has area <cmath>\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|</cmath>. From the problem, this is equal to <math>12</math>. We now solve for y. | ||
+ | |||
+ | <math>\frac{1}{2}|6y - 42| = 12</math> | ||
+ | |||
+ | <math>|6y-42| = 24</math> | ||
+ | |||
+ | <math>6y - 42 = 24</math> OR <math>6y - 42 = -24</math> | ||
+ | |||
+ | <math>6y = 66</math> OR <math>6y = 18</math> | ||
+ | |||
+ | <math>y = 11</math> OR <math>y = 3</math> | ||
+ | |||
+ | However, since, as stated in the problem, <math>y > 7</math>, our only valid solution is <math>\boxed{\textbf{(D)} 11}</math>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==Solution 4== | ||
+ | As in the figure, the triangle is determined by the vectors <math>\begin{bmatrix}-2 \\ y-7\end{bmatrix}</math> and <math>\begin{bmatrix}6\\0\end{bmatrix}</math>. Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that <math>\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24</math>. Expanding the determinants, we find that <math>-6(y-7) = 24</math> or <math>-6(y-7) = -24</math>. Solving each equation individually, we find that <math>y = 3</math> or <math>y = 11</math>. However, the problem states that <math>y > 7</math>, so the only valid solution is <math>\boxed{\textbf{(D)} 11}</math>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] (again!) | ||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
− | https://youtu.be/ | + | https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315 |
~Math-X | ~Math-X | ||
+ | |||
==Video Solution (easy to digest) by Power Solve== | ==Video Solution (easy to digest) by Power Solve== | ||
https://www.youtube.com/watch?v=2UIVXOB4f0o | https://www.youtube.com/watch?v=2UIVXOB4f0o | ||
+ | |||
+ | |||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | ==Video Solution 3 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=RRTxlduaDs8 | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://www.youtube.com/watch?v=-64aBL-lEVg | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/ktzijuZtDas&t=1063 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2024|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Revision as of 16:05, 17 June 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution (easy to digest) by Power Solve
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Interstigation
- 12 See Also
Problem
The coordinates of are
,
, and
, with
. The area of
is 12. What is the value of
?
Solution 1
The triangle has base which means its height satisfies
This means that
so the answer is
Solution 2
Label point
as the point at which
. We now have
, where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of
are
and
. The two side lengths of
are
and
Now,
Dividing by gives
so
-Benedict T (countmath1)
Solution 3
By the Shoelace Theorem, has area
. From the problem, this is equal to
. We now solve for y.
OR
OR
OR
However, since, as stated in the problem, , our only valid solution is
.
~ cxsmi
Solution 4
As in the figure, the triangle is determined by the vectors and
. Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that
. Expanding the determinants, we find that
or
. Solving each equation individually, we find that
or
. However, the problem states that
, so the only valid solution is
.
~ cxsmi (again!)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315
~Math-X
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=RRTxlduaDs8
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=-64aBL-lEVg
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1063
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.