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Difference between revisions of "2021 AIME II Problems/Problem 5"

m (→‎Solution: 4sqrt84 -> 2sqrt84)
(Added in Solution 2.)
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For positive real numbers <math>s</math>, let <math>\tau(s)</math> denote the set of all obtuse triangles that have area <math>s</math> and two sides with lengths <math>4</math> and <math>10</math>. The set of all <math>s</math> for which <math>\tau(s)</math> is nonempty, but all triangles in <math>\tau(s)</math> are congruent, is an interval <math>[a,b)</math>. Find <math>a^2+b^2</math>.
 
For positive real numbers <math>s</math>, let <math>\tau(s)</math> denote the set of all obtuse triangles that have area <math>s</math> and two sides with lengths <math>4</math> and <math>10</math>. The set of all <math>s</math> for which <math>\tau(s)</math> is nonempty, but all triangles in <math>\tau(s)</math> are congruent, is an interval <math>[a,b)</math>. Find <math>a^2+b^2</math>.
  
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==Solution==
+
==Solution 1==
 
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and 14, exclusive. The area of these triangles are from 0 (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between 0 and 20.  
 
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and 14, exclusive. The area of these triangles are from 0 (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between 0 and 20.  
 
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>.
 
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>.
  
 
~ARCTICTURN
 
~ARCTICTURN
 +
 +
==Solution 2 (Casework: Detailed Explanation of Solution 1)==
 +
Every obtuse triangle must satisfy <i><b>both</b></i> of the following:
 +
* <b>Triangle Inequality Theorem</b>
 +
* <b>Pythagorean Inequality</b>
 +
 +
<b>Solution in progress. No edit please</b>
 +
 +
~MRENTHUSIASM
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2021|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:17, 23 March 2021

Problem

For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$.

Solution 1

We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and $\sqrt{84}$ exclusive, and the larger bound is between $\sqrt{116}$ and 14, exclusive. The area of these triangles are from 0 (straight line) to $2\sqrt{84}$ on the first "small bound" and the larger bound is between 0 and 20. $0 < s < 2\sqrt{84}$ is our first equation, and $0 < s < 20$ is our 2nd equation. Therefore, the area is between $\sqrt{336}$ and $\sqrt{400}$, so our final answer is $\boxed{736}$.

~ARCTICTURN

Solution 2 (Casework: Detailed Explanation of Solution 1)

Every obtuse triangle must satisfy both of the following:

  • Triangle Inequality Theorem
  • Pythagorean Inequality

Solution in progress. No edit please

~MRENTHUSIASM

See also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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