Difference between revisions of "2014 AMC 10A Problems/Problem 14"
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<math> \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 </math> | <math> \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
==Solution 1== | ==Solution 1== | ||
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label("$a$",(0,-5),W); | label("$a$",(0,-5),W); | ||
label("$a$",(3,4),NW); | label("$a$",(3,4),NW); | ||
− | + | label("$P$",(0,10),SW); | |
+ | label("$Q$",(0,-10),NW); | ||
+ | // wanted to import graph and use xaxis/yaxis but w/e | ||
+ | label("$x$",(9,0),E); | ||
+ | label("$y$",(0,13),N); | ||
</asy> | </asy> | ||
Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>. | Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>. | ||
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Since both lines contain the point <math>(6,8)</math>, we can plug this into the two equations to obtain <cmath>8=6m+b</cmath> and <cmath>8=-6\frac{1}{m}-b</cmath> | Since both lines contain the point <math>(6,8)</math>, we can plug this into the two equations to obtain <cmath>8=6m+b</cmath> and <cmath>8=-6\frac{1}{m}-b</cmath> | ||
− | Adding the two equations gives <cmath>16=6m+\frac{-6}{m}</cmath> Multiplying by <math>m</math> gives <cmath>16m=6m^2-6</cmath> <cmath>\implies 6m^2-16m-6=0</cmath> <cmath>\implies 3m^2-8m-3=0</cmath> Factoring gives <cmath>( | + | Adding the two equations gives <cmath>16=6m+\frac{-6}{m}</cmath> Multiplying by <math>m</math> gives <cmath>16m=6m^2-6</cmath> <cmath>\implies 6m^2-16m-6=0</cmath> <cmath>\implies 3m^2-8m-3=0</cmath> Factoring gives <cmath>(3m+1)(m-3)=0</cmath> |
+ | |||
+ | Plugging <math>m=3</math> into one of our original equations, we obtain <cmath>8=6(3)+b</cmath> <cmath>\implies b=8-6(3)=-10</cmath> | ||
+ | |||
+ | Since <math>\bigtriangleup APQ</math> has hypotenuse <math>2|b|=20</math> and the altitude to the hypotenuse is equal to the the x-coordinate of point <math>A</math>, or 6, the area of <math>\bigtriangleup APQ</math> is equal to <cmath>\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\ 60}</cmath> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Like Solution 2 but solving directly for intercepts (b): | ||
+ | |||
+ | 1. Solve for m using: <math>8=6m+b</math> | ||
+ | |||
+ | <cmath>m=\frac{8-b}{6}</cmath> | ||
+ | |||
+ | 2. Substitute into the other equation: | ||
+ | |||
+ | <cmath>8=-6\cdot(\frac{1}{\frac{8-b}{6}})-b</cmath> | ||
− | + | Flip the inverse: | |
− | + | <cmath>8=-6\cdot(\frac{6}{8-b})-b</cmath> | |
+ | |||
+ | Multiply <math>6</math>'s: | ||
+ | |||
+ | <cmath>8=-(\frac{36}{8-b})-b</cmath> | ||
+ | |||
+ | |||
+ | 3. Multiply through by <math>8-b</math> (Watch distributing minus!) | ||
+ | |||
+ | <cmath>64-8b=-36-8b+b^2</cmath> | ||
+ | |||
+ | 4. Add <math>36</math> to both sides, and cancel <math>-8b</math> by adding to both sides: | ||
+ | |||
+ | <cmath>100=b^2</cmath> | ||
+ | |||
+ | <math>b=10</math> (or <math>-10</math>) | ||
+ | |||
+ | The rest is as above. | ||
+ | |||
+ | |||
+ | ==Solution 4(Heron's Formula)== | ||
+ | |||
+ | Since their sum is <math>0</math>, let the y intercepts be P<math>(0,a)</math> and Q<math>(0,-a)</math>. The slope of <math>AP</math> is <math>\frac{8-a}{6}</math>. The slope of AQ is <math>\frac{8+a}{6}</math>. Since multiplying the slopes of perpendicular lines yields a product of <math>-1</math>, we have <math>\frac{64-a^2}{36}=-1</math>, which results in <math>a^2=100</math>. We can use either the positive or negative solution because if we choose <math>10</math>, then the other y-intercept is <math>-10</math>; but if we choose <math>-10</math>, then the other y-intercept is <math>10</math>. For simplicity, we choose that <math>a=10</math> in this solution. | ||
− | + | Now we have a triangle APQ with points A<math>(6,8)</math>, P<math>(0,10)</math>, and Q<math>(0,-10)</math>. By the Pythagorean theorem, we have that <math>AP=\sqrt{6^2+2^2}=2\sqrt{10}</math>, and that <math>AQ=\sqrt{6^2+18^2}=6\sqrt{10}</math>. <math>PQ</math> is obviously <math>10--10=20</math> since they have the same <math>x</math> coordinate. Now using Heron's formula, we have | |
+ | <math>\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{D}</math>. | ||
− | + | ~smartninja2000 | |
==See Also== | ==See Also== |
Revision as of 11:53, 7 September 2021
Problem
The -intercepts, and , of two perpendicular lines intersecting at the point have a sum of zero. What is the area of ?
Solution 1
Note that if the -intercepts have a sum of , the distance from the origin to each of the intercepts must be the same. Call this distance . Since the , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is , this means , and the length of the hypotenuse is . Since the -coordinate of is the same as the altitude to the hypotenuse, .
Solution 2
We can let the two lines be This is because the lines are perpendicular, hence the and , and the sum of the y-intercepts is equal to 0, hence the .
Since both lines contain the point , we can plug this into the two equations to obtain and
Adding the two equations gives Multiplying by gives Factoring gives
Plugging into one of our original equations, we obtain
Since has hypotenuse and the altitude to the hypotenuse is equal to the the x-coordinate of point , or 6, the area of is equal to
Solution 3
Like Solution 2 but solving directly for intercepts (b):
1. Solve for m using:
2. Substitute into the other equation:
Flip the inverse:
Multiply 's:
3. Multiply through by (Watch distributing minus!)
4. Add to both sides, and cancel by adding to both sides:
(or )
The rest is as above.
Solution 4(Heron's Formula)
Since their sum is , let the y intercepts be P and Q. The slope of is . The slope of AQ is . Since multiplying the slopes of perpendicular lines yields a product of , we have , which results in . We can use either the positive or negative solution because if we choose , then the other y-intercept is ; but if we choose , then the other y-intercept is . For simplicity, we choose that in this solution.
Now we have a triangle APQ with points A, P, and Q. By the Pythagorean theorem, we have that , and that . is obviously since they have the same coordinate. Now using Heron's formula, we have .
~smartninja2000
See Also
2014 AMC 10A (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.