2021 AIME II Problems/Problem 5

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)

If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:

Triangle Inequality Theorem: $a+b>c$

Pythagorean Inequality Theorem: $a^2+b^2<c^2$

For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:

Case (1): The longest side has length $\boldsymbol{10,}$ so $\boldsymbol{0<x<10.}$

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

By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\sqrt{84}.$

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 $K\in\left(0,2\sqrt{84}\right).$

Case (2): The longest side has length $\boldsymbol{x,}$ so $\boldsymbol{x\geq10.}$

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

By the Pythagorean Inequality Theorem, we have $4^2+10^2<x^2,$ from which $x>\sqrt{116}.$

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

It is possible for non-congruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\left(0,2\sqrt{84}\right)$ or $\left(0,20\right).$ Taking the exclusive disjunction, the set of all such $s$ is $[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$

~MRENTHUSIASM

Solution 3

We have the diagram below.

$[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("A",(0,0),SW); label("B",(1,2*sqrt(3)),N); label("C",(10,0),SE); label("$\theta$",(0,0),NE); label("$\alpha$",(1,2*sqrt(3)),SSE); label("$4$",(0,0)--(1,2*sqrt(3)),WNW); label("$10$",(0,0)--(10,0),S); [/asy]$

We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .

If angle $\theta$ is obtuse, then we have that $s \in (0,20)$ . This is because $s=20$ is attained at $\theta = 90^{\circ}$ , and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$ . This can be observed from $s=\frac{1}{2}(4)(10)\sin\theta$ by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$ .

Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$ . This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$ , yileding $s=4\sqrt{21}$ . Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.

$\angle{ACB}$ cannot be obtuse since $AC>AB$ .

Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$ , the desired range is $s\in [4\sqrt{21},20)$ giving $a^2+b^2=\boxed{736}\Box$

Solution 4

Note: Archimedes15 Solution which I added an answer here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most $1$ . Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \rightarrow x < 2\sqrt{21}$ . Using the same logic as the other case, the area is at most $4\sqrt{21}$ . Square and add $4\sqrt{21}$ and $20$ to get the right answer $a^2+b^2= \boxed{736}\Box$

Solution 5 (Diagrams)

For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$

As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints.

The region in which $\angle B$ is obtuse is determined by construction.

The region in which $\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem.

... DIAGRAM READY SOON ...

For any fixed value of $s,$ we need obtuse $\triangle ABC$ to be unique, or there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on $\overarc{C_1C_2}$ (Error compiling LaTeX. Unknown error_msg)

Let the brackets denote areas.

... DIAGRAM READY SOON ...

WILL BE BACK SOON. NO EDIT PLEASE.

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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