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

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==Problem==
 
==Problem==
  
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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 $\triangle
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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
  
 
==Solution 1==
 
==Solution 1==
  
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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>.  
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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>.  
  
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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>.
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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==
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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 <math>50</math>, 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>.
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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}$.
  
 
==See Also==
 
==See Also==

Revision as of 13:47, 4 May 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

==Solution 1==

Notice that whatever point we pick for$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$\boxed{\textbf{(A) }0}$.

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

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