Difference between revisions of "2015 AMC 12A Problems/Problem 20"
(→Operation Descartes) |
(→Solution 4 (When You're Running Out of Time)) |
||
Line 68: | Line 68: | ||
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>. | 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 | + | ==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, | 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> | <math>2a+b=18</math> and <math>9*(9-a)^2(9-b) = 9*4^2*1</math> |
Revision as of 20:57, 8 December 2020
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 |