2019 AMC 10B Problems/Problem 10: Difference between revisions

Latest revision as of 15:45, 11 November 2025

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 (A bit tedious)

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. ~hashbrown2009

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 (kinda bashy quadratics)

Since $AB$ has a length of $10$ , the height must be $10$ as well for the area to be $50$ , so we can drop an altitude from $C$ that has length $10$ , and call the foot of the altitude on $AB$ as $D$ . Let $AD$ have length $x$ , so $BD$ must have length $10-x$ . From the Pythagorean Theorem, $AC$ has length $\sqrt{x^2+100}$ and $BC$ has length $\sqrt{(10-x)^2+100}$ . For the perimeter to be $50$ , $10+\sqrt{x^2+100}+\sqrt{(10-x)^2+100}=50$ , so $40-\sqrt{x^2+100}=\sqrt{(10-x)^2+100}$ . Squaring both sides and simplifying, $1600-80\sqrt{x^2+100}=-20x+100$ . We can rearrange this and divide all sides by $10$ to get $150+2x=8\sqrt{x^2+100}$ . Squaring both sides yet again, rearranging, and simplifying some more, we get the quadratic $3x^2-30x-805=0$ . So, from the quadratic formula, $x = \frac{30+\sqrt{900+4\cdot3\cdot805}}{6}$ or $x = \frac{30-\sqrt{900+4\cdot3\cdot805}}{6}$ . Immediately we can rule out the negative solution (it's negative because the square root of the discriminant is greater than $30$ , and we'll use this later too), as $x$ must be a positive number since it is a length. However, for the positive solution, we see that the discriminant is greater than $900$ , so the square root is greater than $30$ , so the entire expression when divided by $6$ for $x$ is greater than $10$ , but this is not possible because $x$ should be less than $10$ . Therefore, the number of possibilities is $\boxed{\textbf{(A) }0}$ .

~vaishnav

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

@@ Line 3: / Line 3: @@
 ==Problem==
-In a given plane, points <math>A</math> and <math>B</math> are <math>10</math> units apart. How many points <math>C</math> are there in the plane such that the perimeter of <math>\triangle ABC</math> is <math>50</math> units and the area of <math>\triangle
+In a given plane, points <math>A</math> and <math>B</math> are <math>10</math> units apart. How many points <math>C</math> are there in the plane such that the perimeter of <math>\triangle ABC</math> is <math>50</math> units and the area of <math>\triangle ABC</math> is <math>100</math> square units?
+<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}</math>
 ==Solution 1==
-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>(0,10)<math>, since for any other combination of points, we can just rotate the plane to make them </math>(0,0)<math> and </math>(0,10)<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>.
+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>.
-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>.
+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==
-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}$.
+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 (A bit tedious)==
+We have:
+. Area = <math>100</math>
+. Perimeter = <math>50</math>
+. Semiperimeter <math>s = 50 \div 2 = 25</math>
+We let:
+. <math>z = \overline{AB} = 10</math>
+. <math>x = \overline{AC}</math>
+. <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.
+~hashbrown2009
+==Solution 4 (graphing)==
+First, let's assume that A and B are <imath>(-5,0)</imath> and <imath>(5,0)</imath> respectively. The graph of "the perimeter is <imath>50</imath>" means that <imath>\overline{AC}+\overline{BC}=50-10=40</imath>. So this is the graph of an ellipse (memorize that!). Now let the endpoints of the major axis be <imath>(-x,0)</imath> and <imath>(x,0)</imath>. Then <imath>(x-5)+(x+5)=40</imath> and <imath>x=20</imath>. So the <imath>2</imath> endpoints of the major axis are <imath>(-20,0)</imath> and <imath>(20,0)</imath>. We can also figure out the endpoints of the minor axis must have a y-coordinate less than <imath>20</imath>. It is actually <imath>\sqrt{395}</imath>.
+Now, we consider "the area is <imath>100</imath>". Since the base has length <imath>10</imath>, then the height must have length <imath>20</imath>. So the graph of "the area is 100" is <imath>2</imath> lines, one at <imath>y=20</imath> and the other at <imath>y=-20</imath>. However, this graph does NOT intersect the ellipse, as <imath>\sqrt{395} < 20</imath>. So, there are no intersections and thus no solutions, so the answer is <imath>\boxed{\textbf{(A) }0}</imath>.
+~Yrock
+==Solution 5 (kinda bashy quadratics)==
+Since <imath>AB</imath> has a length of <imath>10</imath>, the height must be <imath>10</imath> as well for the area to be <imath>50</imath>, so we can drop an altitude from <imath>C</imath> that has length <imath>10</imath>, and call the foot of the altitude on <imath>AB</imath> as <imath>D</imath>. Let <imath>AD</imath> have length <imath>x</imath>, so <imath>BD</imath> must have length <imath>10-x</imath>. From the Pythagorean Theorem, <imath>AC</imath> has length <imath>\sqrt{x^2+100}</imath> and <imath>BC</imath> has length <imath>\sqrt{(10-x)^2+100}</imath>. For the perimeter to be <imath>50</imath>, <imath>10+\sqrt{x^2+100}+\sqrt{(10-x)^2+100}=50</imath>, so <imath>40-\sqrt{x^2+100}=\sqrt{(10-x)^2+100}</imath>. Squaring both sides and simplifying, <imath>1600-80\sqrt{x^2+100}=-20x+100</imath>. We can rearrange this and divide all sides by <imath>10</imath> to get <imath>150+2x=8\sqrt{x^2+100}</imath>. Squaring both sides yet again, rearranging, and simplifying some more, we get the quadratic <imath>3x^2-30x-805=0</imath>. So, from the quadratic formula, <imath>x = \frac{30+\sqrt{900+4\cdot3\cdot805}}{6}</imath> or <imath>x = \frac{30-\sqrt{900+4\cdot3\cdot805}}{6}</imath>. Immediately we can rule out the negative solution (it's negative because the square root of the discriminant is greater than <imath>30</imath>, and we'll use this later too), as <imath>x</imath> must be a positive number since it is a length. However, for the positive solution, we see that the discriminant is greater than <imath>900</imath>, so the square root is greater than <imath>30</imath>, so the entire expression when divided by <imath>6</imath> for <imath>x</imath> is greater than <imath>10</imath>, but this is not possible because <imath>x</imath> should be less than <imath>10</imath>. Therefore, the number of possibilities is <imath>\boxed{\textbf{(A) }0}</imath>.
+~vaishnav
+==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
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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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2019 AMC 10B Problems/Problem 10: Difference between revisions

Latest revision as of 15:45, 11 November 2025

Contents

Problem

Solution 1

Solution 2

Solution 3 (A bit tedious)

Solution 4 (graphing)

Solution 5 (kinda bashy quadratics)

Video Solution

Video Solution

See Also