Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
 
(33 intermediate revisions by 18 users not shown)
Line 7: Line 7:
 
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}</math>
 
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}</math>
  
−
==Solution==
+
==Solution 1==
  
−
Notice that whatever point we pick for C, AB will be the base of the triangle. WLOG, points A and B are (0,0) and (0,10) [notice that for any other combination of points, we can just rotate the plane to be the same thing]. When we pick point C, we have to make sure the y value of C is 20, because that's the only way the area of the triangle can be 100.  
+
Notice that whatever point we pick for <math>C</math>, <math>AB</math> will be the base of the triangle. Without loss of generality, let points <math>A</math> and <math>B</math> be <math>(0,0)</math> and <math>(10,0)</math>, since for any other combination of points, we can just rotate the plane to make them <math>(0,0)</math> and <math>(10,0)</math> under a new coordinate system. When we pick point <math>C</math>, we have to make sure that its <math>y</math>-coordinate is <math>\pm20</math>, because that's the only way the area of the triangle can be <math>100</math>.  
  
−
We figure that the one thing we need to test to see if there is such a triangle is when the perimeter is minimized, and the value of C is (x, 20). Thus, we put C in the middle, so point C is (5, 20). We can easily see that AC and BC will both be <math>\sqrt{20^2+5^2} \Rightarrow \sqrt{425}</math>. The perimeter of this minimized triangle is <math>2\sqrt{425} + 10</math>, which is larger than 50. Since the minimized perimeter is greater than 50, there is no triangle that satisfies the condition, giving us <math>\boxed{\textbf{(A) }0}</math>.
+
Now when the perimeter is minimized, by symmetry, we put <math>C</math> in the middle, at <math>(5, 20)</math>. We can easily see that <math>AC</math> and <math>BC</math> will both be <math>\sqrt{20^2+5^2} = \sqrt{425}</math>. The perimeter of this minimal triangle is <math>2\sqrt{425} + 10</math>, which is larger than <math>50</math>. Since the minimum perimeter is greater than <math>50</math>, there is no triangle that satisfies the condition, giving us <math>\boxed{\textbf{(A) }0}</math>.
 +
 
 +
~IronicNinja
  
 
==Solution 2==
 
==Solution 2==
−
WLOG let <math>AB</math> be a horizontal segment of length 10. Now, realize <math>C</math> has to lie on one of the lines parallel to <math>AB</math> and vertically 20 units away from it. But 10+20+20 is already 50, and this doesn't form a triangle. Otherwise, WLOG <math>AC<20</math>. Dropping altitude <math>CD</math> we have a right triangle <math>ACD</math> with hypotenuse <math>AC<20</math> and leg <math>CD=20</math> which is clearly impossible, giving us <math>\boxed{\textbf{(A) }0}</math>.
+
Without loss of generality, let <math>AB</math> be a horizontal segment of length <math>10</math>. Now realize that <math>C</math> has to lie on one of the lines parallel to <math>AB</math> and vertically <math>20</math> units away from it. But <math>10+20+20</math> is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, <math>AC<20</math>. Dropping altitude <math>CD</math>, we have a right triangle <math>ACD</math> with hypotenuse <math>AC<20</math> and leg <math>CD=20</math>, which is clearly impossible, again giving the answer as <math>\boxed{\textbf{(A) }0}</math>.
 +
 
 +
==Solution 3==
 +
We have:
 +
 
 +
1. Area = <math>100</math>
 +
 
 +
2. Perimeter = <math>50</math>
 +
 
 +
3. Semiperimeter <math>s = 50 \div 2 = 25</math>
 +
 
 +
We let:
 +
 
 +
1. <math>z = \overline{AB} = 10</math>
 +
 
 +
2. <math>x = \overline{AC}</math>
 +
 
 +
3. <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>\left((-40)^2 - 4 \times 1 \times 401 \frac{2}{3}\right) < 0</math>. Thus, there are no real solutions.
 +
 
 +
==Solution 4 (graphing)==
 +
 
 +
First, let's assume that A and B are <math>(-5,0)</math> and <math>(5,0)</math> respectively. The graph of "the perimeter is <math>50</math>" means that <math>\overline{AC}+\overline{BC}=50-10=40</math>. So this is the graph of an ellipse (memorize that!). Now let the endpoints of the major axis be <math>(-x,0)</math> and <math>(x,0)</math>. Then <math>(x-5)+(x+5)=40</math> and <math>x=20</math>. So the <math>2</math> endpoints of the major axis are <math>(-20,0)</math> and <math>(20,0)</math>. We can also figure out the endpoints of the minor axis must have a y-coordinate less than <math>20</math>. It is actually <math>\sqrt{395}</math>.
 +
 
 +
Now, we consider "the area is <math>100</math>". Since the base has length <math>10</math>, then the height must have length <math>20</math>. So the graph of "the area is 100" is <math>2</math> lines, one at <math>y=20</math> and the other at <math>y=-20</math>. However, this graph does NOT intersect the ellipse, as <math>\sqrt{395} < 20</math>. So, there are no intersections and thus no solutions, so the answer is <math>\boxed{\textbf{(A) }0}</math>.
 +
 
 +
~Yrock
 +
 
 +
 
 +
==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==
 
==See Also==
Line 21: Line 74:
 
{{AMC12 box|year=2019|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2019|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}
−
SUB2PEWDS
 

Latest revision as of 08:37, 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


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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_10&oldid=234863"