Difference between revisions of "2015 AMC 12A Problems/Problem 20"
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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, we have <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>\textbf{(A)}</math>. |
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+ | ===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(\frac12b)(\frac12b)(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>\textbf{(A)}</math>. | ||
===Solution 2=== | ===Solution 2=== |
Revision as of 21:44, 9 April 2018
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
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 b 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 b.
We get 67.375 which is too high, so we know that .
The answer is .
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 .
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 |