Difference between revisions of "2019 AMC 10B Problems/Problem 10"

Revision as of 14:12, 18 January 2020

The following problem is from both the 2019 AMC 10B #10 and 2019 AMC 12B #6, so both problems redirect to this page.

Problem

In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$

Solution 1

Notice that whatever point we pick for $C$ , $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(0,10)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(0,10)$ under a new coordinate system. When we pick point $C$ , we have to make sure that its $y$ -coordinate is $\pm20$ , because that's the only way the area of the triangle can be $100$ .

Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$ . We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$ . The perimeter of this minimal triangle is $2\sqrt{425} + 10$ , which is larger than $50$ . Since the minimum perimeter is greater than $50$ , there is no triangle that satisfies the condition, giving us $\boxed{\textbf{(A) }0}$ .

~IronicNinja

Solution 2

Without loss of generality, let $AB$ be a horizontal segment of length $10$ . Now realize that $C$ has to lie on one of the lines parallel to $AB$ and vertically $20$ units away from it. But $10+20+20$ is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, $AC<20$ . Dropping altitude $CD$ , we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ , which is clearly impossible, again giving the answer as $\boxed{\textbf{(A) }0}$ .

Solution 3

Area = $100$ , perimeter = $50$ , semiperimeter $s = 50/2 = 25$ , $z = AB = 10$ , $x = AC$ and $y = 50-10-x = 40-x$ .

Using the generic formula for triangle area using semiperimeter $s$ and sides $x$ , $y$ , and $z$ , area = $\sqrt{(s)(s-x)(s-y)(s-z)}$ . (Heron's formula)

$100 = \sqrt{(25)(25-10)(25-x)(25-(40-x))} = \sqrt{(375)(25-x)(x-15)}$ .

Square both sides, divide by $375$ and expand the polynomial to get $35x - x^2 - 375 = 80/3$ .

$x^2 -10x + (375+80/3) = 0$ and the discriminant is $((-35)^2 - 4*1*401.\overline{6}) < 0$ so no real solutions.

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 27: / Line 27: @@
-Square both sides, divide by <math>375</math> and expand the polynomial to get <math>10x - x^2 - 375 = 80/3</math>.
+Square both sides, divide by <math>375</math> and expand the polynomial to get <math>35x - x^2 - 375 = 80/3</math>.
-<math>x^2 -10x + (375+80/3) = 0</math> and the discriminant is <math>((-10)^2 - 4*1*401.\overline{6}) < 0</math> so no real solutions.
+<math>x^2 -10x + (375+80/3) = 0</math> and the discriminant is <math>((-35)^2 - 4*1*401.\overline{6}) < 0</math> so no real solutions.
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