Difference between revisions of "2021 AIME II Problems/Problem 5"
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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 4 and 10 as the sides, so we know that the | + | 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 | + | <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 | ||
− | ==Solution 2 (Casework | + | ==Solution 2 (Inequalities and Casework)== |
If <math>a,b,</math> and <math>c</math> are the side-lengths of an obtuse triangle with <math>a\leq b\leq c,</math> then both of the following must be satisfied: | If <math>a,b,</math> and <math>c</math> are the side-lengths of an obtuse triangle with <math>a\leq b\leq c,</math> then both of the following must be satisfied: | ||
− | * < | + | * <i>Triangle Inequality Theorem:</i> <math>a+b>c</math> |
− | * < | + | * <i>Pythagorean Inequality Theorem:</i> <math>a^2+b^2<c^2</math> |
For one such obtuse triangle, let <math>4,10,</math> and <math>x</math> be its side-lengths and <math>K</math> be its area. We apply casework to its longest side: | For one such obtuse triangle, let <math>4,10,</math> and <math>x</math> be its side-lengths and <math>K</math> be its area. We apply casework to its longest side: | ||
Line 39: | Line 39: | ||
<b>Answer</b> | <b>Answer</b> | ||
− | It is possible for | + | 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== | ||
Line 78: | Line 79: | ||
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 ( | + | ==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)); | ||
+ | |||
+ | 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: | ||
+ | <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> | ||
+ | <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)); | ||
− | Let the brackets denote areas | + | Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); |
− | <ul style="list-style-type: | + | Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); |
− | <li>If <math>C=C_1,</math> then <math>[ABC]</math> will be minimized. By the same base and height and the Inscribed Angle Theorem, we have | + | 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: | ||
+ | <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 | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
[ABC]&=[ABD] \\ | [ABC]&=[ABD] \\ | ||
Line 103: | Line 160: | ||
&=2\sqrt{84}. | &=2\sqrt{84}. | ||
\end{align*}</cmath></li><p> | \end{align*}</cmath></li><p> | ||
− | <li>If <math>C=C_2,</math> then <math>[ABC]</math> will be maximized. For this right triangle, we have | + | <li>If <math>C=C_2,</math> then <math>[ABC]</math> will be maximized (unattainable). For this right triangle, we have |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
[ABC]&=\frac12\cdot AB\cdot BC \\ | [ABC]&=\frac12\cdot AB\cdot BC \\ | ||
Line 110: | Line 167: | ||
\end{align*}</cmath></li><p> | \end{align*}</cmath></li><p> | ||
</ul> | </ul> | ||
− | Finally, | + | 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) | ||
+ | |||
+ | == 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. | ||
+ | |||
+ | Therefore, <math>s</math> is in the range <math>\left[ 4 \sqrt{21}, 20\right)</math>, so answer is <math>\left(4 \sqrt{21}\right)^{2} + 20^{2} = \boxed{736}</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 8== | ||
+ | If <math>4</math> and <math>10</math> are the shortest sides and <math>\angle C</math> is the included angle, then the area is <cmath>\frac{4\cdot10\cdot\sin\angle C}{2} = 20\sin\angle C.</cmath> | ||
+ | Because <math>0\leq\sin\angle C\leq1</math>, the maximum value of <math>20\sin\angle C</math> is <math>20</math>, so <math>s\leq20</math>. | ||
+ | |||
+ | If <math>4</math> is a shortest side and <math>10</math> is the longest side, the length of the other short side is <math>4\cos\angle C+2\sqrt{4\cos^2 \angle C+21}</math> by law of cosines, and the area is <math>2\left(4\cos\angle C+2\sqrt{4\cos^2\angle C+21}\right)\sqrt{1-\cos\angle C}</math>. Because <math>-1\le \cos\angle C\le 0</math>, this is minimized if <math>\cos\angle C=0</math>, where <math>s=4\sqrt{21}</math>. | ||
+ | |||
+ | So, the answer is <math>20^2+\left(4\sqrt{21}\right)^2=\boxed{736}</math>. | ||
+ | |||
+ | ~ryanbear | ||
− | + | ==Video Solution by Interstigation== | |
+ | https://youtu.be/ODokTEt3EVA | ||
− | + | ~Interstigation | |
− | ==See | + | ==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}} |
Latest revision as of 00:43, 2 January 2023
Contents
Problem
For positive real numbers , let
denote the set of all obtuse triangles that have area
and two sides with lengths
and
. The set of all
for which
is nonempty, but all triangles in
are congruent, is an interval
. Find
.
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 and
as the sides, so we know that the
rd side is between
and
, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the
rd side. So the triangles' sides are between
and
exclusive, and the larger bound is between
and
, exclusive. The area of these triangles are from
(straight line) to
on the first "small bound" and the larger bound is between
and
.
is our first equation, and
is our
nd equation. Therefore, the area is between
and
, so our final answer is
.
~ARCTICTURN
Solution 2 (Inequalities and Casework)
If and
are the side-lengths of an obtuse triangle with
then both of the following must be satisfied:
- Triangle Inequality Theorem:
- Pythagorean Inequality Theorem:
For one such obtuse triangle, let and
be its side-lengths and
be its area. We apply casework to its longest side:
Case (1): The longest side has length so
By the Triangle Inequality Theorem, we have from which
By the Pythagorean Inequality Theorem, we have from which
Taking the intersection produces for this case.
At the obtuse triangle degenerates into a straight line with area
at
the obtuse triangle degenerates into a right triangle with area
Together, we obtain
or
Case (2): The longest side has length so
By the Triangle Inequality Theorem, we have from which
By the Pythagorean Inequality Theorem, we have from which
Taking the intersection produces for this case.
At the obtuse triangle degenerates into a straight line with area
at
the obtuse triangle degenerates into a right triangle with area
Together, we obtain
or
Answer
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers are in exactly one of
or
By the exclusive disjunction, the set of all such
is
from which
~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 that they yield .
If angle is obtuse, then we have that
. This is because
is attained at
, and the area of the triangle is strictly decreasing as
increases beyond
. This can be observed from
by noting that
is decreasing in
.
Then, we note that if is obtuse, we have
. This is because we get
when
, yileding
. Then,
is decreasing as
increases by the same argument as before.
cannot be obtuse since
.
Now we have the intervals and
for the cases where
and
are obtuse, respectively. We are looking for the
that are in exactly one of these intervals, and because
, the desired range is
giving
Solution 4
Note: Archimedes15 Solution which I added an answer
here are two cases. Either the and
are around an obtuse angle or the
and
are around an acute triangle. If they are around the obtuse angle, the area of that triangle is
as we have
and
is at most
. Note that for the other case, the side lengths around the obtuse angle must be
and
where we have
. Using the same logic as the other case, the area is at most
. Square and add
and
to get the right answer
Solution 5 (Circles)
For we fix
and
Without the loss of generality, we consider
on only one side of
As shown below, all locations for at which
is an obtuse triangle are indicated in red, excluding the endpoints.
Note that:
- The region in which
is obtuse is determined by construction.
- The region in which
is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
For any fixed value of the height from
is fixed. We need obtuse
to be unique, so there can only be one possible location for
As shown below, all possible locations for
are on minor arc
including
but excluding
Let the brackets denote areas:
- If
then
will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have
- If
then
will be maximized (unattainable). For this right triangle, we have
Finally, the set of all such is
from which
~MRENTHUSIASM (credit given to Snowfan)
Solution 6
Let a triangle in be
, where
and
. We will proceed with two cases:
Case 1: is obtuse. If
is obtuse, then, if we imagine
as the base of our triangle, the height can be anything in the range
; therefore, the area of the triangle will fall in the range of
.
Case 2: is obtuse. Then, if we imagine
as the base of our triangle, the height can be anything in the range
. Therefore, the area of the triangle will fall in the range of
.
If , there will exist two types of triangles in
- one type with
obtuse; the other type with
obtuse. If
, as we just found,
cannot be obtuse, so therefore, there is only one type of triangle - the one in which
is obtuse. Also, if
, no triangle exists with lengths
and
. Therefore,
is in the range
, so our answer is
.
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 and
when there is exactly one such obtuse triangle. In the diagram,
The largest area of triangle with sides and
is
for a right triangle with legs
and
(
).
The diagram shows triangles with equal heights. The yellow triangle has the longest side
the blue triangle
has the longest side
If
then
the area is equal to
In the interval, the blue triangle
is acute-angled, the yellow triangle
is obtuse-angled. Their heights and areas are equal. The condition is met.
If the area is less than both triangles are obtuse, not equal, so the condition is not met.
Therefore, is in the range
, so answer is
vladimir.shelomovskii@gmail.com, vvsss
Solution 8
If and
are the shortest sides and
is the included angle, then the area is
Because
, the maximum value of
is
, so
.
If is a shortest side and
is the longest side, the length of the other short side is
by law of cosines, and the area is
. Because
, this is minimized if
, where
.
So, the answer is .
~ryanbear
Video Solution by Interstigation
~Interstigation
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.