Difference between revisions of "2015 AMC 12A Problems/Problem 20"
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<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math> | <math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math> | ||
− | + | ==Solution 1 == | |
− | |||
The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18. | The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18. | ||
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Thus <math>12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math>, so <math>48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}</math>. | Thus <math>12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math>, so <math>48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}</math>. | ||
− | We square and divide 36 from both sides to obtain <math>64 = b^2 (9 - b)</math>, so <math>b^3 - 9b^2 + 64 = 0</math>. Since we know <math>b = 8</math> is a solution, we divide by <math>b - 8</math> to get the other solution. Thus, <math>b^2 - b - 8 = 0</math>, so <math>b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.</math> The answer is <math>\textbf{(A)}</math>. | + | We square and divide 36 from both sides to obtain <math>64 = b^2 (9 - b)</math>, so <math>b^3 - 9b^2 + 64 = 0</math>. Since we know <math>b = 8</math> is a solution, we divide by <math>b - 8</math> to get the other solution. Thus, <math>b^2 - b - 8 = 0</math>, so <math>b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.</math> The answer is <math>\boxed{\textbf{(A) }3}</math>. |
===Solution 1.1=== | ===Solution 1.1=== | ||
− | The area is <math>12</math>, the semiperimeter is <math>9</math>, and <math>a = 9 - \frac12b</math>. Using Heron's formula, <math>\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9-b)} = 12</math>. Squaring both sides and simplifying, we have <math>-b^3+9b-64=0</math>. Since we know <math>b = 8</math> is a solution, we divide by <math>b - 8</math> to get the other solution. Thus, <math>b^2 - b - 8 = 0</math>, so <math>b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.</math> The answer is <math>\boxed{\textbf{(A)}}</math>. | + | The area is <math>12</math>, the semiperimeter is <math>9</math>, and <math>a = 9 - \frac12b</math>. Using Heron's formula, <math>\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9-b)} = 12</math>. Squaring both sides and simplifying, we have <math>-b^3+9b^2-64=0</math>. Since we know <math>b = 8</math> is a solution, we divide by <math>b - 8</math> to get the other solution. Thus, <math>b^2 - b - 8 = 0</math>, so <math>b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.</math> The answer is <math>\boxed{\textbf{(A) }3}</math>. |
− | + | ==Solution 2== | |
Triangle <math>T</math>, being isosceles, has an area of <math>\frac{1}{2}(8)\sqrt{5^2-4^2}=12</math> and a perimeter of <math>5+5+8=18</math>. | Triangle <math>T</math>, being isosceles, has an area of <math>\frac{1}{2}(8)\sqrt{5^2-4^2}=12</math> and a perimeter of <math>5+5+8=18</math>. | ||
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Utilizing the quadratic formula gives | Utilizing the quadratic formula gives | ||
<cmath>b=\frac{1\pm\sqrt{33}}{2}</cmath> | <cmath>b=\frac{1\pm\sqrt{33}}{2}</cmath> | ||
− | We clearly must pick the positive solution. Note that <math>5<\sqrt{33}<6</math>, and so <math>{3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}</math>, which clearly gives an answer of <math>\ | + | We clearly must pick the positive solution. Note that <math>5<\sqrt{33}<6</math>, and so <math>{3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}</math>, which clearly gives an answer of <math>\boxed{\textbf{(A) }3}</math>, as desired. |
− | + | ==Solution 3== | |
Triangle T has perimeter <math>5 + 5 + 8 = 18</math> so <math>18 = 2a + b</math>. | Triangle T has perimeter <math>5 + 5 + 8 = 18</math> so <math>18 = 2a + b</math>. | ||
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We get <math>67.375</math> which is too high, so we know that <math>b < 3.5</math>. | We get <math>67.375</math> which is too high, so we know that <math>b < 3.5</math>. | ||
− | + | Thus the answer is <math>\boxed{\textbf{(A) }3}</math> | |
− | == | + | ==Solution 4 (Operation Descartes)== |
For this new triangle, say its legs have length <math>d</math> and the base length <math>2c</math>. To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that <math>c+d=9</math> and <math>c*\sqrt{d^2-c^2}=12</math>. It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!! | For this new triangle, say its legs have length <math>d</math> and the base length <math>2c</math>. To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that <math>c+d=9</math> and <math>c*\sqrt{d^2-c^2}=12</math>. It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!! | ||
− | Now, modify the square-root equation with <math>d=9-c</math>; you get <math>c^2*(81-18c)=144</math>, so <math>-18c^3+81c^2=144</math>. Divide by <math>-9</math> to get <math>2c^3-9c^2+16=0</math>. Obviously, <math>c=4</math> is a root as established by triangle <math>T</math>! So, use synthetic division to obtain <math>2c^2-c-4=0</math>, upon which <math>c=\frac{1+\sqrt{33}}{4}</math>, which is closest to <math>\frac{3}{2}</math> (as opposed to <math>2</math>). That's enough to confirm that the answer has to be <math>\textbf{A}</math>. | + | Now, modify the square-root equation with <math>d=9-c</math>; you get <math>c^2*(81-18c)=144</math>, so <math>-18c^3+81c^2=144</math>. Divide by <math>-9</math> to get <math>2c^3-9c^2+16=0</math>. Obviously, <math>c=4</math> is a root as established by triangle <math>T</math>! So, use synthetic division to obtain <math>2c^2-c-4=0</math>, upon which <math>c=\frac{1+\sqrt{33}}{4}</math>, which is closest to <math>\frac{3}{2}</math> (as opposed to <math>2</math>). That's enough to confirm that the answer has to be <math>\boxed{\textbf{(A) }3}</math>. |
+ | |||
+ | ==Solution 5 (When You're Running Out of Time)== | ||
+ | Since triangles <math>T'</math> and <math>T</math> have the same area and the same perimeter, | ||
+ | <math>2a+b=18</math> and <math>9*(9-a)^2(9-b) = 9*4^2*1</math> | ||
+ | By trying each answer choice, it is clear that the answer is <math>\boxed{\textbf{(A) }3}</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}} |
Latest revision as of 21:32, 7 June 2021
Contents
Problem
Isosceles triangles and are not congruent but have the same area and the same perimeter. The sides of have lengths , , and , while those of have lengths , , and . Which of the following numbers is closest to ?
Solution 1
The area of is and the perimeter is 18.
The area of is and the perimeter is .
Thus , so .
Thus , so .
We square and divide 36 from both sides to obtain , so . Since we know is a solution, we divide by to get the other solution. Thus, , so The answer is .
Solution 1.1
The area is , the semiperimeter is , and . Using Heron's formula, . Squaring both sides and simplifying, we have . Since we know is a solution, we divide by to get the other solution. Thus, , so The answer is .
Solution 2
Triangle , being isosceles, has an area of and a perimeter of . Triangle similarly has an area of and .
Now we apply our computational fortitude.
Plug in to obtain Plug in to obtain We know that is a valid solution by . Factoring out , we obtain Utilizing the quadratic formula gives We clearly must pick the positive solution. Note that , and so , which clearly gives an answer of , as desired.
Solution 3
Triangle T has perimeter so .
Using Heron's, we get .
We know that from above so we plug that in, and we also know that then .
We plug in 3 for in the LHS, and we get 54 which is too low. We plug in 4 for in the LHS, and we get 80 which is too high. We now know that is some number between 3 and 4.
If , then we would round up to 4, but if , then we would round down to 3. So let us plug in 3.5 for .
We get which is too high, so we know that .
Thus the answer is
Solution 4 (Operation Descartes)
For this new triangle, say its legs have length and the base length . To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that and . It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!
Now, modify the square-root equation with ; you get , so . Divide by to get . Obviously, is a root as established by triangle ! So, use synthetic division to obtain , upon which , which is closest to (as opposed to ). That's enough to confirm that the answer has to be .
Solution 5 (When You're Running Out of Time)
Since triangles and have the same area and the same perimeter, and By trying each answer choice, it is clear that the answer is .
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |