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Difference between revisions of "2014 AMC 10A Problems/Problem 14"

(Created page with "==Problem== The <math>y</math>-intercepts, <math>P</math> and <math>Q</math>, of two perpendicular lines intersecting at the point <math>A(6,8)</math> have a sum of zero. What i...")
 
 
(25 intermediate revisions by 15 users not shown)
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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==
 +
<asy>//Needs refining (hmm I think it's fine --bestwillcui1)
 +
size(12cm);
 +
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));
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for(int i=-2;i<=8;i+=1)
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  draw((i,-12)--(i,12),grey);
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for(int j=-12;j<=12;j+=1)
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  draw((-2,j)--(8,j),grey);
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draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis
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draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis
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dot((0,0));
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dot((6,8));
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draw((-2,10.66667)--(8,7.33333),Arrows);
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draw((7.33333,12)--(-0.66667,-12),Arrows);
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draw((6,8)--(0,8));
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draw((6,8)--(0,0));
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draw(rightanglemark((0,10),(6,8),(0,-10),20));
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label("$A$",(6,8),NE);
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label("$a$", (0,5),W);
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label("$a$",(0,-5),W);
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label("$a$",(3,4),NW);
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label("$P$",(0,10),SW);
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label("$Q$",(0,-10),NW);
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// wanted to import graph and use xaxis/yaxis but w/e
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label("$x$",(9,0),E);
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label("$y$",(0,13),N);
 +
</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>.
 +
 +
==Solution 2==
 +
 +
We can let the two lines be <cmath>y=mx+b</cmath> <cmath>y=-\frac{1}{m}x-b</cmath> This is because the lines are perpendicular, hence the <math>m</math> and <math>-\frac{1}{m}</math>, and the sum of the y-intercepts is equal to 0, hence the <math>b, -b</math>.
 +
 +
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>(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>
 +
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<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
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 +
==Solution 5 (point-slope)==
 +
Using point-slope form, the first line has the equation <cmath>y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8</cmath>
 +
The second line has the equation <cmath>y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8</cmath>
 +
At the y-intercept, the value of the x-coordinate is <math>0</math>, hence: the first equation is <math>y=-6m+8</math> and the second is <math>y=\frac{6}{m}+8</math>. Since the y-intercepts sum to <math>0</math>, they are opposites, so:
 +
<cmath>-6m+8=-\left(\frac{6}{m}+8\right)=-\frac{6}{m}-8</cmath>
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<cmath>6m-\frac{6}{m}=16</cmath>
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Multiply both sides by m:
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<math>6m^{2}-6=16m \longrightarrow 3m^{2}-8m-3=0</math>. The solution to this quadratic, using the quadratic formula, is:
 +
<math>\frac{8\pm\sqrt{64-4\left(3\right)\left(-3\right)}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}</math>
 +
This yields <math>m=-\frac{1}{3}</math> and <math>m=3</math>. Plugging <math>m=3</math> into the second equation, we get <math>y=\frac{6}{3}+8=10</math>. Plugging <math>m=-\frac{1}{3}</math> into the first equation, we get <math>y=-10</math> So the base is <math>20</math> and the height is <math>6</math>, the area is <math>60 \Longrightarrow \boxed{\textbf{(D) } 60}</math>.
 +
 +
~JH. L
 +
 +
==Solution 6 (Geometry only)==
 +
(Not to scale)
 +
<asy>
 +
unitsize(36);
 +
pair A = (0,0);
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pair B = (5,1);
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pair C = (13/5,13);
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pair D = C-B;
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pair E = (26/5,0);
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pair F = (0,26);
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pair G = intersectionpoints(C--D,A--F)[0];
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draw(A--E--F--A--B--C--D--A,linewidth(1));
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label(A,scale(2)*"A",dir(-135));
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label(E,scale(2)*"E",dir(-45));
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label(F,scale(2)*"F",dir(90));
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label(B,scale(2)*"B",dir(45));
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label(C,scale(2)*"C",dir(45));
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label(D,scale(2)*"D",dir(180));
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label(G,scale(2)*"G",dir(135));
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</asy>
 +
Long solution:
 +
By rotating the right triangle, we get the figure shown where <cmath>CD\perp EF</cmath> and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also. By similar triangles, <cmath>\frac{AD}{FC}=\frac{DG}{CG}~\text{so}~DG=\frac{8}{3}~\text{and}~CG=\frac{10}{3}</cmath>
 +
Then, by more similar triangles,
 +
<cmath>\frac{CG}{FG}=\frac{EA}{FA}=\frac{1}{3}~\text{so}~AF=3AE</cmath>
 +
Then <math>AE=2\sqrt{10},~AE=6\sqrt{10}</math>, and the area is <cmath>\boxed{(D)~60}</cmath>
 +
Short solution:
 +
By rotating the right triangle, we get the figure shown where <cmath>CD\perp EF</cmath> and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also, so CF=20.
 +
<cmath>A=\frac{bh}{2}=\frac{(EF)(AB)}{2}=\frac{(20)(6)}{2}=\frac{120}{2}=\boxed{(\textbf{D})~60}</cmath>
 +
[[User:Afly|Afly]] ([[User talk:Afly|talk]])
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=AJdRK51xvos
 +
~Mathematical Dexterity
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2014|ab=A|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Latest revision as of 20:58, 3 August 2024

Problem

The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?

$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution 1

[asy]//Needs refining (hmm I think it's fine --bestwillcui1) size(12cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1) draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1) draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis dot((0,0)); dot((6,8)); draw((-2,10.66667)--(8,7.33333),Arrows); draw((7.33333,12)--(-0.66667,-12),Arrows); draw((6,8)--(0,8)); draw((6,8)--(0,0)); draw(rightanglemark((0,10),(6,8),(0,-10),20)); label("$A$",(6,8),NE); label("$a$", (0,5),W); label("$a$",(0,-5),W); 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] Note that if the $y$-intercepts have a sum of $0$, the distance from the origin to each of the intercepts must be the same. Call this distance $a$. Since the $\angle PAQ = 90^\circ$, 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 $\sqrt{6^2+8^2} = 10$, this means $a=10$, and the length of the hypotenuse is $2a = 20$. Since the $x$-coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}$.

Solution 2

We can let the two lines be \[y=mx+b\] \[y=-\frac{1}{m}x-b\] This is because the lines are perpendicular, hence the $m$ and $-\frac{1}{m}$, and the sum of the y-intercepts is equal to 0, hence the $b, -b$.

Since both lines contain the point $(6,8)$, we can plug this into the two equations to obtain \[8=6m+b\] and \[8=-6\frac{1}{m}-b\]

Adding the two equations gives \[16=6m+\frac{-6}{m}\] Multiplying by $m$ gives \[16m=6m^2-6\] \[\implies 6m^2-16m-6=0\] \[\implies 3m^2-8m-3=0\] Factoring gives \[(3m+1)(m-3)=0\]

Plugging $m=3$ into one of our original equations, we obtain \[8=6(3)+b\] \[\implies b=8-6(3)=-10\]

Since $\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altitude to the hypotenuse is equal to the the x-coordinate of point $A$, or 6, the area of $\bigtriangleup APQ$ is equal to \[\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\ 60}\]

Solution 3

Like Solution 2 but solving directly for intercepts (b):

1. Solve for m using: $8=6m+b$

\[m=\frac{8-b}{6}\]

2. Substitute into the other equation:

\[8=-6\cdot(\frac{1}{\frac{8-b}{6}})-b\]

Flip the inverse:

\[8=-6\cdot(\frac{6}{8-b})-b\]

Multiply $6$'s:

\[8=-(\frac{36}{8-b})-b\]


3. Multiply through by $8-b$ (Watch distributing minus!)

\[64-8b=-36-8b+b^2\]

4. Add $36$ to both sides, and cancel $-8b$ by adding to both sides:

\[100=b^2\]

$b=10$ (or $-10$)

The rest is as above.


Solution 4(Heron's Formula)

Since their sum is $0$, let the y intercepts be P$(0,a)$ and Q$(0,-a)$. The slope of $AP$ is $\frac{8-a}{6}$. The slope of AQ is $\frac{8+a}{6}$. Since multiplying the slopes of perpendicular lines yields a product of $-1$, we have $\frac{64-a^2}{36}=-1$, which results in $a^2=100$. We can use either the positive or negative solution because if we choose $10$, then the other y-intercept is $-10$; but if we choose $-10$, then the other y-intercept is $10$. For simplicity, we choose that $a=10$ in this solution.

Now we have a triangle APQ with points A$(6,8)$, P$(0,10)$, and Q$(0,-10)$. By the Pythagorean theorem, we have that $AP=\sqrt{6^2+2^2}=2\sqrt{10}$, and that $AQ=\sqrt{6^2+18^2}=6\sqrt{10}$. $PQ$ is obviously $10-(-10)=20$ since they have the same $x$ coordinate. Now using Heron's formula, we have $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{D}$.

~smartninja2000

Solution 5 (point-slope)

Using point-slope form, the first line has the equation \[y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8\] The second line has the equation \[y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8\] At the y-intercept, the value of the x-coordinate is $0$, hence: the first equation is $y=-6m+8$ and the second is $y=\frac{6}{m}+8$. Since the y-intercepts sum to $0$, they are opposites, so: \[-6m+8=-\left(\frac{6}{m}+8\right)=-\frac{6}{m}-8\] \[6m-\frac{6}{m}=16\] Multiply both sides by m: $6m^{2}-6=16m \longrightarrow 3m^{2}-8m-3=0$. The solution to this quadratic, using the quadratic formula, is: $\frac{8\pm\sqrt{64-4\left(3\right)\left(-3\right)}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}$ This yields $m=-\frac{1}{3}$ and $m=3$. Plugging $m=3$ into the second equation, we get $y=\frac{6}{3}+8=10$. Plugging $m=-\frac{1}{3}$ into the first equation, we get $y=-10$ So the base is $20$ and the height is $6$, the area is $60 \Longrightarrow \boxed{\textbf{(D) } 60}$.

~JH. L

Solution 6 (Geometry only)

(Not to scale) [asy] unitsize(36); pair A = (0,0); pair B = (5,1); pair C = (13/5,13); pair D = C-B; pair E = (26/5,0); pair F = (0,26); pair G = intersectionpoints(C--D,A--F)[0]; draw(A--E--F--A--B--C--D--A,linewidth(1)); label(A,scale(2)*"A",dir(-135)); label(E,scale(2)*"E",dir(-45)); label(F,scale(2)*"F",dir(90)); label(B,scale(2)*"B",dir(45)); label(C,scale(2)*"C",dir(45)); label(D,scale(2)*"D",dir(180)); label(G,scale(2)*"G",dir(135)); [/asy] Long solution: By rotating the right triangle, we get the figure shown where \[CD\perp EF\] and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also. By similar triangles, \[\frac{AD}{FC}=\frac{DG}{CG}~\text{so}~DG=\frac{8}{3}~\text{and}~CG=\frac{10}{3}\] Then, by more similar triangles, \[\frac{CG}{FG}=\frac{EA}{FA}=\frac{1}{3}~\text{so}~AF=3AE\] Then $AE=2\sqrt{10},~AE=6\sqrt{10}$, and the area is \[\boxed{(D)~60}\] Short solution: By rotating the right triangle, we get the figure shown where \[CD\perp EF\] and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also, so CF=20. \[A=\frac{bh}{2}=\frac{(EF)(AB)}{2}=\frac{(20)(6)}{2}=\frac{120}{2}=\boxed{(\textbf{D})~60}\] Afly (talk)

Video Solution

https://www.youtube.com/watch?v=AJdRK51xvos ~Mathematical Dexterity

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

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