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

Revision as of 17:37, 29 August 2021

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

We have:

1. Area = $100$

2. Perimeter = $50$

3. Semiperimeter $s = 50 \div 2 = 25$

We let:

1. $z = \overline{AB} = 10$

2. $x = \overline{AC}$

3. $y = 50-10-x = 40-x$ .

Heron's formula states that for real numbers $x$ , $y$ , $z$ , and semiperimeter $s$ , the area is $\sqrt{(s)(s-x)(s-y)(s-z)}$ .

Plugging numbers in, we have $100 = \sqrt{(25)(25-10)(25-x)(25-(40-x))} = \sqrt{(375)(25-x)(x-15)}$ .

Square both sides, divide by $375$ and expand the polynomial to get $40x - x^2 - 375 = \frac{80}{3}$ .

$x^2 - 40x + \left(375 + \frac{80}{3}\right) = 0$ and the discriminant is $((-40)^2 - 4 \times 1 \times 401 \frac{2}{3}) < 0$ . Thus, there are no real solutions.

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

https://youtu.be/INvRdwQzC-w

~savannahsolver

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 19: / Line 19: @@
 ==Solution 3==
-Area = <math>100</math>, perimeter = <math>50</math>, semiperimeter <math>s = 50/2 = 25</math>, <math>z = AB = 10</math>, <math>x = AC</math> and <math>y = 50-10-x = 40-x</math>.
+We have:
+. Area = <math>100</math>
-Using the generic formula for triangle area using semiperimeter <math>s</math> and sides <math>x</math>, <math>y</math>, and <math>z</math>, area = <math>\sqrt{(s)(s-x)(s-y)(s-z)}</math>. (Heron's formula)
+. Perimeter = <math>50</math>
-<math>100 = \sqrt{(25)(25-10)(25-x)(25-(40-x))} = \sqrt{(375)(25-x)(x-15)}</math>.
+. Semiperimeter <math>s = 50 \div 2 = 25</math>
+We let:
-Square both sides, divide by <math>375</math> and expand the polynomial to get <math>40x - x^2 - 375 = 80/3</math>.
+. <math>z = \overline{AB} = 10</math>
+. <math>x = \overline{AC}</math>
-<math>x^2 -40x + (375+80/3) = 0</math> and the discriminant is <math>((-40)^2 - 4*1*401.\overline{6}) < 0</math> so no real solutions.
+. <math>y = 50-10-x = 40-x</math>.
+Heron's formula states that for real numbers <math>x</math>, <math>y</math>, <math>z</math>, and semiperimeter <math>s</math>, the area is <math>\sqrt{(s)(s-x)(s-y)(s-z)}</math>.
+Plugging numbers in, we have <math>100 = \sqrt{(25)(25-10)(25-x)(25-(40-x))} = \sqrt{(375)(25-x)(x-15)}</math>.
+Square both sides, divide by <math>375</math> and expand the polynomial to get <math>40x - x^2 - 375 = \frac{80}{3}</math>.
+<math>x^2 - 40x + \left(375 + \frac{80}{3}\right) = 0</math> and the discriminant is <math>((-40)^2 - 4 \times 1 \times 401 \frac{2}{3}) < 0</math>. Thus, there are no real solutions.
 ==Video Solution==

Page

Toolbox

Search