Difference between revisions of "2015 AMC 12A Problems/Problem 20"
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<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>\fbox{A}</math>, as desired. | 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>\fbox{A}</math>, as desired. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Triangle T has perimeter <math>5 + 5 + 8 = 18</math> so <math>18 = 2a + b</math>. | ||
+ | |||
+ | Using Heron's, we get <math>\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})(\frac{d}{2})^2(\frac{2a-b}{2})}</math>. | ||
+ | |||
+ | We know that <math>2a + b = 18</math> from above so we plug that in, and we also know that then <math>2a - b = 18 - 2b</math>. | ||
+ | |||
+ | <math>12 = 3\frac{b}{2}\sqrt{9-b}</math> | ||
+ | |||
+ | <math>64 = 9b^2 - b^3</math> | ||
+ | |||
+ | We plug in 3 for <math>b</math> in the LHS, and we get 54 which is too low. We plug in 4 for <math>b</math> in the LHS, and we get 80 which is too high. We now know that b is some number between 3 and 4. | ||
+ | |||
+ | If <math>b \geq 3.5</math>, then we would round up to 4, but if <math>b < 3.5</math>, 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 <math>b < 3.5</math>. | ||
+ | |||
+ | The answer is <math>3</math>. <math>\textbf{(A)}</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}} |
Revision as of 11:53, 31 January 2016
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 . This factors as . Because clearly but , we have 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 .
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 |