Difference between revisions of "2015 AMC 12A Problems/Problem 20"
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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>. | ||
− | Using Heron's, we get <math>\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})(\frac{b}{2})^2(\frac{2a-b}{2})}</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>. | 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 = | + | <math>12 = \frac{3b}{2}\sqrt{9-b}</math> |
<math>64 = 9b^2 - b^3</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. | + | 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 b. | + | 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 67.375 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>. |
− | The answer is <math> | + | The answer is <math>\boxed{\textbf{(A) }3}</math> |
===Operation Descartes=== | ===Operation Descartes=== |
Revision as of 13:32, 1 February 2019
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
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
.
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 |