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 "2021 AIME II Problems/Problem 5"

(→‎Solution 2 (Casework: Detailed Explanation of Solution 1))
(→‎Solution 2 (Casework: Detailed Explanation of Solution 1))
Line 25: Line 25:
 
Taking the intersection produces <math>6<x<\sqrt{84}</math> for this case.
 
Taking the intersection produces <math>6<x<\sqrt{84}</math> for this case.
  
−
At <math>x=6,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{84},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.</math> Together, we obtain <math>0<k<2\sqrt{84}.</math>
+
At <math>x=6,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{84},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.</math> Together, we obtain <math>0<k<2\sqrt{84},</math> or <math>k\in\left(0,\sqrt{84}\right).</math>
  
 
<b>Case (2): The longest side has length <math>\boldsymbol{x.}</math></b>
 
<b>Case (2): The longest side has length <math>\boldsymbol{x.}</math></b>
Line 35: Line 35:
 
Taking the intersection produces <math>\sqrt{116}<x<14</math> for this case.
 
Taking the intersection produces <math>\sqrt{116}<x<14</math> for this case.
  
−
At ..., the obtuse triangle degenerates into a straight line with area ...; at ..., the obtuse triangle degenerates into a right triangle with area .... Together, we obtain ....
+
At <math>x=14,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{116},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot10=20.</math> Together, we obtain <math>0<k<20,</math> or <math>k\in\left(0,20\right).</math>
  
 
<b>Answer</b>
 
<b>Answer</b>

Revision as of 15:12, 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: If $a,b,$ and $c$ are the side-lengths of a triangle with $a\leq b\leq c,$ then $a+b>c.$
  • Pythagorean Inequality Theorem: If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b<c,$ then $a^2+b^2<c^2.$

For one such obtuse triangle, let $4,10,$ and $x$ be the side-lengths and $k$ be the area. We will use casework on the longest side:

Case (1): The longest side has length $\boldsymbol{10.}$

By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$

By the Pythagorean Inequality Theorem, we get $4^2+x^2<10^2,$ so that $x<\sqrt{84}.$

Taking the intersection produces $6<x<\sqrt{84}$ for this case.

At $x=6,$ the obtuse triangle degenerates into a straight line with area $k=0;$ at $x=\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.$ Together, we obtain $0<k<2\sqrt{84},$ or $k\in\left(0,\sqrt{84}\right).$

Case (2): The longest side has length $\boldsymbol{x.}$

By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$

By the Pythagorean Inequality Theorem, we get $4^2+10^2<x^2,$ so that $x>\sqrt{116}.$

Taking the intersection produces $\sqrt{116}<x<14$ for this case.

At $x=14,$ the obtuse triangle degenerates into a straight line with area $k=0;$ at $x=\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $k=\frac12\cdot4\cdot10=20.$ Together, we obtain $0<k<20,$ or $k\in\left(0,20\right).$

Answer

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

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=2021_AIME_II_Problems/Problem_5&oldid=150020"