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)
Line 11: Line 11:
 
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 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.  
  
−
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{A) 0}</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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:38, 16 February 2019

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

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.

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 $\sqrt{20^2+5^2} \Rightarrow \sqrt{425}$. The perimeter of this minimized triangle is $2\sqrt{425} + 10$, which is larger than 50. Since the minimized perimeter is greater than 50, there is no triangle that satisfies the condition, giving us $\boxed{\textbf{(A) }0}$.

Solution 2

WLOG let $AB$ be a horizontal segment of length 10. Now, realize $C$ 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, WLOG $AC<20$. Dropping altitude $CD$ we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ which is clearly impossible, giving us $\boxed{A) 0}$.

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

SUB2PEWDS

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_10&oldid=103121"