Difference between revisions of "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 (Inequalities and Casework)

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

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

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 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 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 from which $x>\sqrt{116}.$

Taking the intersection produces $\sqrt{116} 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 or $K\in\left(0,20\right).$

Answer

It is possible for noncongruent 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.

For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$

Let the brackets denote areas:

• If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have \begin{align*} [ABC]&=[ABD] \\ &=\frac12\cdot BD\cdot DA \\ &=\frac12\cdot BD\cdot \sqrt{AB^2-BD^2} \\ &=\frac12\cdot 4\cdot \sqrt{10^2-4^2} \\ &=2\sqrt{84}. \end{align*}
• If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have \begin{align*} [ABC]&=\frac12\cdot AB\cdot BC \\ &=\frac12\cdot 10\cdot 4 \\ &=20. \end{align*}

Finally, we get $$s\in[a,b)=\left[2\sqrt{84},20\right),$$ from which $a^2+b^2=\boxed{736}.$

~MRENTHUSIASM (credit given to Snowfan)

See Also

 2021 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 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.

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