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== | + | ==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. | ||
− | The area of <math>T</math> is <math>\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math> and the perimeter is <math>2a + b</math>. | + | The area of <math>T'</math> is <math>\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math> and the perimeter is <math>2a + b</math>. |
− | Thus <math>2a + b = 18</math>, so <math>2a = 18 - | + | Thus <math>2a + b = 18</math>, so <math>2a = 18 - b</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>. | 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>. | + | 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=== | ||
+ | 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> similarly has an area of <math>\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12</math> and <math>2a+b=18</math>. | ||
+ | |||
+ | Now we apply our computational fortitude. | ||
+ | |||
+ | <cmath>\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12</cmath> | ||
+ | <cmath>(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=24</cmath> | ||
+ | <cmath>(b)\sqrt{4a^2-b^2}=48</cmath> | ||
+ | <cmath>b^2(4a^2-b^2)=48^2</cmath> | ||
+ | <cmath>b^2(2a+b)(2a-b)=48^2</cmath> | ||
+ | Plug in <math>2a+b=18</math> to obtain | ||
+ | <cmath>18b^2(2a-b)=48^2</cmath> | ||
+ | <cmath>b^2(2a-b)=128</cmath> | ||
+ | Plug in <math>2a=18-b</math> to obtain | ||
+ | <cmath>b^2(18-2b)=128</cmath> | ||
+ | <cmath>2b^3-18b^2+128=0</cmath> | ||
+ | <cmath>b^3-9b^2+64=0</cmath> | ||
+ | We know that <math>b=8</math> is a valid solution by <math>T</math>. Factoring out <math>b-8</math>, we obtain | ||
+ | <cmath>(b-8)(b^2-b-8)=0 \Rightarrow b^2-b-8=0</cmath> | ||
+ | Utilizing the quadratic formula gives | ||
+ | <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>\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>. | ||
+ | |||
+ | Using Heron's, we get <math>\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})\left(\frac{b}{2}\right)^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 = \frac{3b}{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 <math>b</math> 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 <math>b</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!! | ||
+ | |||
+ | 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 == | ||
+ | {{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}} |
Latest revision as of 22:32, 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 |