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Difference between revisions of "2019 AMC 10B Problems/Problem 10"
Line 53: Line 53:
  
 
~Yrock
 
~Yrock
 +
 +
==Solution 5==
 +
 +
==Diagram==
 +
<asy>
 +
size(200);
 +
defaultpen(linewidth(0.4)+fontsize(10));
 +
pen s = linewidth(0.8)+fontsize(8);
 +
 +
pair A,B,C,D,E0,F,G;
 +
A = origin;
 +
C = (15,0);
 +
B = IP(CR(A,13),CR(C,14));
 +
D = foot(A,C,B);
 +
E0 = foot(D,A,C);
 +
F = OP(CR((A+B)/2,length(B-A)/2), D--E0);
 +
draw(A--C--B--A, black+0.8);
 +
draw(B--F--A--D--E0);
 +
dot("$A$",A,W);
 +
dot("$B$",B,N);
 +
dot("$C$",C,E);
 +
dot("$D$",D,NE);
 +
dot("$E$",E0,S);
 +
dot("$F$",F,E);
 +
draw(rightanglemark(B,D,A,15));
 +
draw(rightanglemark(B,F,A,15));
 +
draw(rightanglemark(D,E0,A,15));
 +
label("$5$",D--B,NE);
 +
label("$9$",D--C,NE);
 +
label(Label("$13$",Rotate(B-A)), B--A);
 +
</asy>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 08:27, 18 November 2024

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 $(10,0)$, since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ 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 $\left((-40)^2 - 4 \times 1 \times 401 \frac{2}{3}\right) < 0$. Thus, there are no real solutions.

Solution 4 (graphing)

First, let's assume that A and B are $(-5,0)$ and $(5,0)$ respectively. The graph of "the perimeter is $50$" means that $\overline{AC}+\overline{BC}=50-10=40$. So this is the graph of an ellipse (memorize that!). Now let the endpoints of the major axis be $(-x,0)$ and $(x,0)$. Then $(x-5)+(x+5)=40$ and $x=20$. So the $2$ endpoints of the major axis are $(-20,0)$ and $(20,0)$. We can also figure out the endpoints of the minor axis must have a y-coordinate less than $20$. It is actually $\sqrt{395}$.

Now, we consider "the area is $100$". Since the base has length $10$, then the height must have length $20$. So the graph of "the area is 100" is $2$ lines, one at $y=20$ and the other at $y=-20$. However, this graph does NOT intersect the ellipse, as $\sqrt{395} < 20$. So, there are no intersections and thus no solutions, so the answer is $\boxed{\textbf{(A) }0}$.

~Yrock

Solution 5

Diagram

[asy] size(200); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,E0,F,G; A = origin; C = (15,0); B = IP(CR(A,13),CR(C,14)); D = foot(A,C,B); E0 = foot(D,A,C); F = OP(CR((A+B)/2,length(B-A)/2), D--E0); draw(A--C--B--A, black+0.8); draw(B--F--A--D--E0); dot("$A$",A,W); dot("$B$",B,N); dot("$C$",C,E); dot("$D$",D,NE); dot("$E$",E0,S); dot("$F$",F,E); draw(rightanglemark(B,D,A,15)); draw(rightanglemark(B,F,A,15)); draw(rightanglemark(D,E0,A,15)); label("$5$",D--B,NE); label("$9$",D--C,NE); label(Label("$13$",Rotate(B-A)), B--A); [/asy]

Video Solution

https://youtu.be/MNVKkjVvBUU

~Education, the Study of Everything

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

https://youtu.be/INvRdwQzC-w

~savannahsolver

See Also

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

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