2021 AIME II Problems/Problem 5

Revision as of 22:49, 30 October 2021 by MRENTHUSIASM (talk | contribs) (Solution 6: Fixed the parentheses issue. Note that for individual cases, the endpoints are exclusive. For the exclusive disjunction, the left endpoint is inclusive, but the right endpoint is exclusive.)

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 $3$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: $a^2+b^2<c^2$

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 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.

[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 (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. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P));  draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red);  dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill);  Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15));  label("$\angle C$ obtuse",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label("$\angle B$ obtuse",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] 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 $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2));  draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed);  dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot("$D$", D, 1.5*dir(75), linewidth(0.8), UnFill); dot("$C_2$", C2, 1.5*N, linewidth(4.5)); dot("$C_1$", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5));  Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] 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, 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 $s \leq 2 \sqrt{84}$, 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 $\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}$.

~ihatemath123

See Also

2021 AIME II (ProblemsAnswer KeyResources)
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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