AIME 2022 II · 第 5 题

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 $3$rd 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 $3$rd 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 $2$nd 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 $\left(0,20\right).$ By 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.

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 $\alpha=90^{\circ}$, yielding $s=4\sqrt{21}$. 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 (Circles)

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.

Note that:

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

2. 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

$$\begin{aligned} [ABC]&=[ABD] \\ &=\frac12\cdot BD\cdot DA \\ &=\frac12\cdot BD\cdot \sqrt{AB^2-BD^2} \\ &=\frac12\cdot 4\cdot \sqrt{10^2-4^2} \\ &=2\sqrt{84}. \end{aligned}$$

• If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have

$$\begin{aligned} [ABC]&=\frac12\cdot AB\cdot BC \\ &=\frac12\cdot 10\cdot 4 \\ &=20. \end{aligned}$$ Finally, the set of all such $s$ is $[a,b)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$

~MRENTHUSIASM (credit given to Snowfan)

Solution 6

Let a triangle in $\tau(s)$ be $ABC$, where $AB = 4$ and $BC = 10$. We will proceed with two cases:

Case 1: $\angle ABC$ is obtuse. If $\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$; therefore, the area of the triangle will fall in the range of $(0, 20)$.

Case 2: $\angle BAC$ is obtuse. Then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $\left(0, \sqrt{10^{2} - 4^{2}}\right)$. Therefore, the area of the triangle will fall in the range of $\left(0, 2 \sqrt{84}\right)$.

If $s < 2 \sqrt{84}$, there will exist two types of triangles in $\tau(s)$ - one type with $\angle ABC$ obtuse; the other type with $\angle BAC$ obtuse. If $s \geq 2 \sqrt{84}$, as we just found, $\angle BAC$ cannot be obtuse, so therefore, there is only one type of triangle - the one in which $\angle ABC$ is obtuse. Also, if $s > 20$, no triangle exists with lengths $4$ and $10$. Therefore, $s$ is in the range $\left[ 2 \sqrt{84}, 20\right)$, so our answer is $\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}$.

Alternatively, refer to Solution 5 for the geometric interpretation.

~ihatemath123

Solution 7

Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides $4$ and $10,$ when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$

The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10$ ($AC\perp AB$).

The diagram shows triangles with equal heights. The yellow triangle $ABC'$ has the longest side $BC',$ the blue triangle $ABC$ has the longest side $AC.$ If $BC\perp AB,$ then$BC = \sqrt {AC^2 – AB^2} = 2 \sqrt{21}$ the area is equal to $4\sqrt{21}.$ In the interval, the blue triangle $ABC$ is acute-angled, the yellow triangle $ABC'$ is obtuse-angled. Their heights and areas are equal. The condition is met.

If the area is less than $4\sqrt{21},$ both triangles are obtuse, not equal, so the condition is not met.

Therefore, $s$ is in the range $\left[ 4 \sqrt{21}, 20\right)$, so answer is $\left(4 \sqrt{21}\right)^{2} + 20^{2} = \boxed{736}$

vladimir.shelomovskii@gmail.com, vvsss

Solution 8

If $4$ and $10$ are the shortest sides and $\angle C$ is the included angle, then the area is

$$\frac{4\cdot10\cdot\sin\angle C}{2} = 20\sin\angle C.$$ Because $0\leq\sin\angle C\leq1$, the maximum value of $20\sin\angle C$ is $20$, so $s\leq20$.

If $4$ is a shortest side and $10$ is the longest side, the length of the other short side is $4\cos\angle C+2\sqrt{4\cos^2 \angle C+21}$ by law of cosines, and the area is $2\left(4\cos\angle C+2\sqrt{4\cos^2\angle C+21}\right)\sqrt{1-\cos\angle C}$. Because $-1\le \cos\angle C\le 0$, this is minimized if $\cos\angle C=0$, where $s=4\sqrt{21}$.

So, the answer is $20^2+\left(4\sqrt{21}\right)^2=\boxed{736}$.

~ryanbear

Video Solution by Interstigation

https://youtu.be/ODokTEt3EVA

~Interstigation