Difference between revisions of "2021 AIME II Problems/Problem 5"

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(Solution 7)
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==Solution 1==
 
==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 <math>4</math> and <math>10</math> as the sides, so we know that the 3rd side is between <math>6</math> and <math>14</math>, 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 <math>6</math> and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and <math>14</math>, exclusive. The area of these triangles are from <math>0</math> (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between <math>0</math> and <math>20</math>.  
+
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given <math>4</math> and <math>10</math> as the sides, so we know that the <math>3</math>rd side is between <math>6</math> and <math>14</math>, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the <math>3</math>rd side. So the triangles' sides are between <math>6</math> and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and <math>14</math>, exclusive. The area of these triangles are from <math>0</math> (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between <math>0</math> and <math>20</math>.  
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>.
+
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our <math>2</math>nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>.
  
 
~ARCTICTURN
 
~ARCTICTURN
Line 39: Line 39:
 
<b>Answer</b>
 
<b>Answer</b>
  
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers <math>s</math> are in exactly one of <math>\left(0,2\sqrt{84}\right)</math> or <math>\left(0,20\right).</math> Taking the exclusive disjunction, the set of all such <math>s</math> is <cmath>[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),</cmath> from which <math>a^2+b^2=\boxed{736}.</math>
+
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers <math>s</math> are in exactly one of <math>\left(0,2\sqrt{84}\right)</math> or <math>\left(0,20\right).</math> By the exclusive disjunction, the set of all such <math>s</math> is <cmath>[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),</cmath> from which <math>a^2+b^2=\boxed{736}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 52: Line 52:
 
draw((1,2*sqrt(3))--(10,0));
 
draw((1,2*sqrt(3))--(10,0));
 
draw((10,0)--(0,0));
 
draw((10,0)--(0,0));
label("A",(0,0),SW);
+
label("$A$",(0,0),SW);
label("B",(1,2*sqrt(3)),N);
+
label("$B$",(1,2*sqrt(3)),N);
label("C",(10,0),SE);
+
label("$C$",(10,0),SE);
 
label("$\theta$",(0,0),NE);
 
label("$\theta$",(0,0),NE);
 
label("$\alpha$",(1,2*sqrt(3)),SSE);
 
label("$\alpha$",(1,2*sqrt(3)),SSE);
Line 73: Line 73:
 
Now we have the intervals <math>s \in (0,20)</math> and <math>s \in (0,4\sqrt{21})</math> for the cases where <math>\theta</math> and <math>\alpha</math> are obtuse, respectively. We are looking for the <math>s</math> that are in exactly one of these intervals, and because <math>4\sqrt{21}<20</math>, the desired range is
 
Now we have the intervals <math>s \in (0,20)</math> and <math>s \in (0,4\sqrt{21})</math> for the cases where <math>\theta</math> and <math>\alpha</math> are obtuse, respectively. We are looking for the <math>s</math> that are in exactly one of these intervals, and because <math>4\sqrt{21}<20</math>, the desired range is
 
<cmath>s\in [4\sqrt{21},20)</cmath>giving <cmath>a^2+b^2=\boxed{736}\Box</cmath>
 
<cmath>s\in [4\sqrt{21},20)</cmath>giving <cmath>a^2+b^2=\boxed{736}\Box</cmath>
 +
 
==Solution 4==  
 
==Solution 4==  
  
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here are two cases. Either the <math>4</math> and <math>10</math> are around an obtuse angle or the <math>4</math> and <math>10</math> are around an acute triangle. If they are around the obtuse angle, the area of that triangle is <math><20</math> as we have <math>\frac{1}{2} \cdot 40 \cdot \sin{\alpha}</math> and <math>\sin</math> is at most <math>1</math>. Note that for the other case, the side lengths around the obtuse angle must be <math>4</math> and <math>x</math> where we have <math>16+x^2 < 100 \rightarrow x < 2\sqrt{21}</math>. Using the same logic as the other case, the area is at most <math>4\sqrt{21}</math>. Square and add <math>4\sqrt{21}</math> and <math>20</math> to get the right answer <cmath>a^2+b^2= \boxed{736}\Box</cmath>
 
here are two cases. Either the <math>4</math> and <math>10</math> are around an obtuse angle or the <math>4</math> and <math>10</math> are around an acute triangle. If they are around the obtuse angle, the area of that triangle is <math><20</math> as we have <math>\frac{1}{2} \cdot 40 \cdot \sin{\alpha}</math> and <math>\sin</math> is at most <math>1</math>. Note that for the other case, the side lengths around the obtuse angle must be <math>4</math> and <math>x</math> where we have <math>16+x^2 < 100 \rightarrow x < 2\sqrt{21}</math>. Using the same logic as the other case, the area is at most <math>4\sqrt{21}</math>. Square and add <math>4\sqrt{21}</math> and <math>20</math> to get the right answer <cmath>a^2+b^2= \boxed{736}\Box</cmath>
  
==Solution 5 (Diagrams)==
+
==Solution 5 (Circles)==
 
For <math>\triangle ABC,</math> we fix <math>AB=10</math> and <math>BC=4.</math> Without the loss of generality, we consider <math>C</math> on only one side of <math>\overline{AB}.</math>
 
For <math>\triangle ABC,</math> we fix <math>AB=10</math> and <math>BC=4.</math> Without the loss of generality, we consider <math>C</math> on only one side of <math>\overline{AB}.</math>
  
 
As shown below, all locations for <math>C</math> at which <math>\triangle ABC</math> is an obtuse triangle are indicated in red, excluding the endpoints.
 
As shown below, all locations for <math>C</math> at which <math>\triangle ABC</math> 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));
  
* The region in which <math>\angle B</math> is obtuse is determined by construction.
+
draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q);
 +
draw(Arc(B,Q,C2)^^Arc(B,D,P),red);
  
* The region in which <math>\angle C</math> is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
+
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);
  
[[File:2021 AIME II Problem 5 (1).png|center]]
+
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:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>The region in which <math>\angle B</math> is obtuse is determined by construction.</li><p>
 +
  <li>The region in which <math>\angle C</math> is obtuse is determined by the corollaries of the Inscribed Angle Theorem.</li><p>
 +
</ol>
 
For any fixed value of <math>s,</math> the height from <math>C</math> is fixed. We need obtuse <math>\triangle ABC</math> to be unique, so there can only be one possible location for <math>C.</math> As shown below, all possible locations for <math>C</math> are on minor arc <math>\widehat{C_1C_2},</math> including <math>C_1</math> but excluding <math>C_2.</math>
 
For any fixed value of <math>s,</math> the height from <math>C</math> is fixed. We need obtuse <math>\triangle ABC</math> to be unique, so there can only be one possible location for <math>C.</math> As shown below, all possible locations for <math>C</math> are on minor arc <math>\widehat{C_1C_2},</math> including <math>C_1</math> but excluding <math>C_2.</math>
 +
<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);
  
[[File:2021 AIME II Problem 5 (2).png|center]]
+
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:  
 
Let the brackets denote areas:  
<ul style="list-style-type:disc;">
+
<ul style="list-style-type:square;">
 
   <li>If <math>C=C_1,</math> then <math>[ABC]</math> will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have
 
   <li>If <math>C=C_1,</math> then <math>[ABC]</math> will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 110: Line 167:
 
\end{align*}</cmath></li><p>
 
\end{align*}</cmath></li><p>
 
</ul>
 
</ul>
Finally, we get <cmath>s\in[a,b)=\left[2\sqrt{84},20\right),</cmath> from which <math>a^2+b^2=\boxed{736}.</math>
+
Finally, the set of all such <math>s</math> is <math>[a,b)=\left[2\sqrt{84},20\right),</math> from which <math>a^2+b^2=\boxed{736}.</math>
  
 
~MRENTHUSIASM (credit given to Snowfan)
 
~MRENTHUSIASM (credit given to Snowfan)
 +
 +
== Solution 6 ==
 +
Let a triangle in <math>\tau(s)</math> be <math>ABC</math>, where <math>AB = 4</math> and <math>BC = 10</math>. We will proceed with two cases:
 +
 +
Case 1: <math>\angle ABC</math> is obtuse. If <math>\angle ABC</math> is obtuse, then, if we imagine <math>AB</math> as the base of our triangle, the height can be anything in the range <math>(0,10)</math>; therefore, the area of the triangle will fall in the range of <math>(0, 20)</math>.
 +
 +
Case 2: <math>\angle BAC</math> is obtuse. Then, if we imagine <math>AB</math> as the base of our triangle, the height can be anything in the range <math>\left(0, \sqrt{10^{2} - 4^{2}}\right)</math>. Therefore, the area of the triangle will fall in the range of <math>\left(0, 2 \sqrt{84}\right)</math>.
 +
 +
If <math>s < 2 \sqrt{84}</math>, there will exist two types of triangles in <math>\tau(s)</math> - one type with <math>\angle ABC</math> obtuse; the other type with <math>\angle BAC</math> obtuse. If <math>s \geq 2 \sqrt{84}</math>, as we just found, <math>\angle BAC</math> cannot be obtuse, so therefore, there is only one type of triangle - the one in which <math>\angle ABC</math> is obtuse. Also, if <math>s > 20</math>, no triangle exists with lengths <math>4</math> and <math>10</math>. Therefore, <math>s</math> is in the range <math>\left[ 2 \sqrt{84}, 20\right)</math>, so our answer is <math>\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}</math>.
 +
 +
Alternatively, refer to Solution 5 for the geometric interpretation.
 +
 +
~ihatemath123
 +
 +
== Solution 7 ==
 +
[[File:2021 AIME II 5.png|400px|right]]
 +
Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides <math>4</math> and <math>10,</math> when there is exactly one such obtuse triangle. In the diagram, <math>AB = 4, AC = 10.</math>
 +
 +
The largest area of triangle with sides <math>4</math> and <math>10</math> is <math>20</math> for a right triangle with legs <math>4</math> and <math>10</math> (<math>AC\perp AB</math>).
 +
 +
The diagram shows triangles with equal heights. The yellow triangle <math>ABC'</math> has the longest side <math>BC',</math> the blue triangle <math>ABC</math> has the longest side <math>AC.</math>
 +
If <math>BC\perp AB,</math> then<math> BC = \sqrt {AC^2 – AB^2} = 2 \sqrt{21}</math> the area is equal to <math>4\sqrt{21}.</math> In the interval, the blue triangle <math>ABC</math> is acute-angled, the yellow triangle <math>ABC'</math> is obtuse-angled. Their heights and areas are equal. The condition is met.
 +
 +
If the area is less than <math>4\sqrt{21},</math> both triangles are obtuse, not equal, so the condition is not met.
 +
 +
'''Shelomovskii, vvsss,  www.deoma-cmd.ru'''
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ODokTEt3EVA
 +
 +
~Interstigation
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2021|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2021|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:20, 19 July 2022

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 $s < 2 \sqrt{84}$, 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

2021 AIME II 5.png

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.

Shelomovskii, vvsss, www.deoma-cmd.ru

Video Solution by Interstigation

https://youtu.be/ODokTEt3EVA

~Interstigation

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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