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== | ||
Line 26: | Line 27: | ||
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 | // wanted to import graph and use xaxis/yaxis but w/e | ||
label("$x$",(9,0),E); | label("$x$",(9,0),E); | ||
Line 40: | Line 43: | ||
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> | 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 | ||
+ | |||
+ | ==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> | ||
+ | <cmath>6m-\frac{6}{m}=16</cmath> | ||
+ | Multiply both sides by m: | ||
+ | <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); | ||
+ | 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 <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== | ==See Also== |
Latest revision as of 20:58, 3 August 2024
Contents
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
Solution 5 (point-slope)
Using point-slope form, the first line has the equation The second line has the equation At the y-intercept, the value of the x-coordinate is , hence: the first equation is and the second is . Since the y-intercepts sum to , they are opposites, so: Multiply both sides by m: . The solution to this quadratic, using the quadratic formula, is: This yields and . Plugging into the second equation, we get . Plugging into the first equation, we get So the base is and the height is , the area is .
~JH. L
Solution 6 (Geometry only)
(Not to scale) Long solution: By rotating the right triangle, we get the figure shown where 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, Then, by more similar triangles, Then , and the area is Short solution: By rotating the right triangle, we get the figure shown where 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. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.