Difference between revisions of "1989 AIME Problems/Problem 10"
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== Problem == | == Problem == | ||
− | Let <math> | + | Let <math>a</math>, <math>b</math>, <math>c</math> be the three sides of a [[triangle]], and let <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math>, be the angles opposite them. If <math>a^2+b^2=1989c^2</math>, find |
<center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center> | <center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center> | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | We | + | === Solution 1 === |
+ | We draw the [[altitude]] <math>h</math> to <math>c</math>, to get two [[right triangle]]s. | ||
+ | <center><asy> | ||
+ | size(170); | ||
+ | pair A = (0,0), B = (3, 0), C = (1, 4); | ||
+ | pair P = .5*(C + reflect(A,B)*C); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(C--P, dotted); | ||
+ | draw(rightanglemark(C,P, B, 4)); | ||
+ | label("$A$", A, S); | ||
+ | label("$B$", B, S); | ||
+ | label("$C$", C, N); | ||
+ | label("$a$", (B+C)/2, NE); | ||
+ | label("$b$", (A+C)/2, NW); | ||
+ | label("$c$", (A+B)/2, S); | ||
+ | label("$h$", (C+P)/2, E);</asy></center> | ||
− | <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the cotangent. | + | Then <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the [[cotangent]]. |
− | + | Let <math>K</math> be the area of <math>\triangle ABC.</math> Then <math>h=\frac{2K}{c}</math>, so <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}</math>. | |
− | + | By identical logic, we can find similar expressions for the sums of the other two cotangents: | |
+ | <cmath> \begin{align*} | ||
+ | \cot \alpha + \cot \beta &= \frac{c^2}{2K} \\ | ||
+ | \cot \beta + \cot \gamma &= \frac{a^2}{2K} \\ | ||
+ | \cot \gamma + \cot \alpha &= \frac{b^2}{2K}. \end{align*} </cmath> | ||
+ | Adding the last two equations, subtracting the first, and dividing by 2, we get | ||
+ | <cmath> \cot \gamma = \frac{a^2 + b^2 - c^2}{4K}.</cmath> | ||
+ | Therefore | ||
+ | <cmath> \begin{align*} | ||
+ | \frac{\cot \gamma}{\cot \alpha + \cot \beta} &= \frac{(a^2 + b^2 - c^2)/(4K)}{c^2/(2K)} \\ | ||
+ | &= \frac{a^2 + b^2 - c^2}{2c^2} \\ | ||
+ | &= \frac{1989 c^2 - c^2}{2c^2} \\ | ||
+ | &= \frac{1988}{2} = \boxed{994}. \end{align*} </cmath> | ||
− | + | === Solution 2 === | |
− | < | + | By the [[law of cosines]], |
+ | <cmath> \cos \gamma = \frac{a^2 + b^2 - c^2}{2ab}. </cmath> So, by the extended [[law of sines]], | ||
+ | <cmath> \cot \gamma = \frac{\cos \gamma}{\sin \gamma} = \frac{a^2 + b^2 - c^2}{2ab} \cdot \frac{2R}{c} = \frac{R}{abc} (a^2 + b^2 - c^2). </cmath> | ||
+ | Identical logic works for the other two angles in the triangle. So, the cotangent of any angle in the triangle is directly proportional to the sum of the squares of the two adjacent sides, minus the square of the opposite side. Therefore | ||
+ | <cmath> \frac{\cot \gamma}{\cot \alpha + \cot \beta} = \frac{a^2 + b^2 - c^2}{(-a^2 + b^2 + c^2) + (a^2 - b^2 + c^2)} = \frac{a^2 + b^2 - c^2}{2c^2}. </cmath> | ||
+ | We can then finish as in solution 1. | ||
− | + | === Solution 3 === | |
− | <math> | + | We start as in solution 1, though we'll write <math>A</math> instead of <math>K</math> for the area. Now we evaluate the numerator: |
− | + | <cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath> | |
− | + | From the [[Law of Cosines]] and the sine area formula, | |
+ | <cmath>\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\ | ||
+ | \sin{\gamma}&= \frac{2A}{ab}\\ | ||
+ | \cot{\gamma}&= \frac{\cos \gamma}{\sin \gamma} = \frac{1988c^2}{4A} \end{align*}</cmath> | ||
+ | |||
+ | Then <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}</math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | <cmath>\begin{align*} | ||
+ | \cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac {\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}\\ &= \frac {\sin{(\alpha + \beta)}}{\sin{\alpha}\sin{\beta}} = \frac {\sin{\gamma}}{\sin{\alpha}\sin{\beta}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | By the Law of Cosines, | ||
+ | |||
+ | <cmath>a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2</cmath> | ||
+ | |||
+ | Now | ||
+ | |||
+ | <cmath>\begin{align*}\frac {\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac {\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma}} = \frac {\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}} = \frac {ab}{c^2}\cos{\gamma} = \frac {ab}{c^2} \cdot \frac {994c^2}{ab}\\ &= \boxed{994}\end{align*}</cmath> | ||
+ | |||
+ | |||
+ | === Solution 5=== | ||
+ | |||
+ | Use Law of cosines to give us <math>c^2=a^2+b^2-2ab\cos(\gamma)</math> or therefore <math>\cos(\gamma)=\frac{994c^2}{ab}</math>. Next, we are going to put all the sin's in term of <math>\sin(a)</math>. We get <math>\sin(\gamma)=\frac{c\sin(a)}{a}</math>. Therefore, we get <math>\cot(\gamma)=\frac{994c}{b\sin a}</math>. | ||
+ | |||
+ | Next, use Law of Cosines to give us <math>b^2=a^2+c^2-2ac\cos(\beta)</math>. Therefore, <math>\cos(\beta)=\frac{a^2-994c^2}{ac}</math>. Also, <math>\sin(\beta)=\frac{b\sin(a)}{a}</math>. Hence, <math>\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}</math>. | ||
+ | |||
+ | Lastly, <math>\cos(\alpha)=\frac{b^2-994c^2}{bc}</math>. Therefore, we get <math>\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}</math>. | ||
+ | |||
+ | Now, <math>\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}</math>. After using <math>a^2+b^2=1989c^2</math>, we get <math>\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}</math>. | ||
+ | |||
+ | |||
+ | === Solution 6=== | ||
+ | |||
+ | |||
+ | Let <math>\gamma</math> be <math>(180-\alpha-\beta)</math> | ||
+ | |||
+ | <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}</math> | ||
+ | |||
+ | WLOG, assume that <math>a</math> and <math>c</math> are legs of right triangle <math>abc</math> with <math>\beta = 90^o</math> and <math>c=1</math> | ||
+ | |||
+ | By the Pythagorean theorem, we have <math>b^2=a^2+1</math>, and the given <math>a^2+b^2=1989</math>. Solving the equations gives us <math>a=\sqrt{994}</math> and <math>b=\sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>. | ||
+ | |||
+ | Our derived equation equals <math>\tan^2 \alpha</math> as <math>\tan \beta</math> approaches infinity. | ||
+ | Evaluating <math>\tan^2 \alpha</math>, we get <math>\boxed{994}</math>. | ||
+ | |||
+ | |||
+ | === Solution 7=== | ||
+ | |||
+ | As in Solution 1, drop an altitude <math>h</math> to <math>c</math>. Let <math>h</math> meet <math>c</math> at <math>P</math>, and let <math>AP = x, BP = y</math>. | ||
+ | |||
+ | <center><asy> | ||
+ | size(170); | ||
+ | pair A = (0,0), B = (3,0), C = (1,4); | ||
+ | pair P = .5*(C + reflect(A,B)*C); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(C--P, dotted); | ||
+ | draw(rightanglemark(C,P, B , 4)); | ||
+ | label("$A$", A, S); | ||
+ | label("$B$", B, S); | ||
+ | label("$C$", C, N); | ||
+ | label("$P$", P, S); | ||
+ | label("$x$", (A+P)/2, S); | ||
+ | label("$y$", (B+P)/2, S); | ||
+ | label("$a$", (B+C)/2, NE); | ||
+ | label("$b$", (A+C)/2, NW); | ||
+ | label("$c$", (A+B)/2, S); | ||
+ | label("$h$", (C+P)/2, E);</asy></center> | ||
+ | |||
+ | Then, <math>\cot{\alpha} = \frac{1}{\tan{\alpha}} = \frac{x}{h}</math>, <math>\cot{\beta} = \frac{1}{\tan{\beta}} = \frac{y}{h}</math>. We can calculate <math>\cot{\gamma}</math> using the [[tangent addition formula]], after noticing that <math>\cot{\gamma} = \frac{1}{\tan{\gamma}}</math>. So, we find that | ||
+ | \begin{align*} | ||
+ | \cot{\gamma} &= \frac{1}{\tan{\gamma}} \\ | ||
+ | &= \frac{1}{\frac{\frac{x}{h} + \frac{y}{h}}{1 - \frac{xy}{h^2}}} \\ | ||
+ | &= \frac{1}{\frac{(x+y)h}{h^2 - xy}} \\ | ||
+ | &= \frac{h^2 - xy}{(x+y)h}. | ||
+ | \end{align*} | ||
+ | |||
+ | So now we can simplify our original expression: | ||
+ | \begin{align*} | ||
+ | \frac{\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac{\frac{h^2 - xy}{(x+y)h}}{\frac{x + y}{h}} \\ | ||
+ | &= \frac{h^2 - xy}{(x+y)^2}. | ||
+ | \end{align*} | ||
+ | |||
+ | But notice that <math>x+y = c</math>, so this becomes <cmath>\frac{h^2 - xy}{c^2}.</cmath> | ||
+ | Now note that we can use the [[Pythagorean theorem]] to calculate <math>h^2</math>, we get that <cmath>h^2 = \frac{a^2 - y^2 + b^2 - x^2}{2}.</cmath> | ||
+ | So our expression simplifies to <cmath>\frac{1989c^2 - (x+y)^2}{2c^2}</cmath> | ||
+ | since <math>a^2 + b^2 = 1989c^2</math> from the problem and that there is another <math>-\frac{2xy}{2}</math> after the <math>h^2</math> in our expression. Again note that <math>x+y = c</math>, so it again simplifies to <math>\frac{1988c^2}{2c^2}</math>, or <math>\boxed{994}</math>. | ||
+ | |||
+ | ~[[User: Yiyj1|Yiyj1]] | ||
+ | |||
+ | |||
+ | |||
+ | === Solution 8 (Quick and Easy) === | ||
+ | |||
+ | Since no additional information is given, we can assume that triangle ABC is right with the right angle at B. | ||
+ | We can use the [[Pythagorean theorem]] to say <cmath>c^2+a^2=b^2</cmath> | ||
+ | We can now solve for <math>a</math> in terms of <math>c</math> | ||
+ | |||
+ | <cmath>c^2+a^2=1989c^2-a^2</cmath> | ||
+ | <cmath>a^2=994c^2</cmath> | ||
+ | <cmath>a=\sqrt{994}c</cmath> | ||
+ | |||
+ | Using the definition of cotangent | ||
+ | |||
+ | <cmath>cot(A)=\frac{c}{a}=\frac{1}{\sqrt{994}}</cmath> | ||
+ | <cmath>cot(B)=cot(90)=0</cmath> | ||
+ | <cmath>cot(C)=\frac{a}{c}=\sqrt{994}</cmath> | ||
+ | Plugging into our desired expression, we get <math>\boxed{994}</math> | ||
+ | |||
+ | ~[[User: ms0001|ms0001]] | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1989|num-b=9|num-a=11}} | |
− | + | ||
− | + | [[Category:Intermediate Geometry Problems]] | |
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:41, 20 January 2024
Problem
Let ,
,
be the three sides of a triangle, and let
,
,
, be the angles opposite them. If
, find
![$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$](http://latex.artofproblemsolving.com/8/5/8/858c49b569dc76c503dbdbfa8d0b78fd2c04fe1d.png)
Contents
Solution
Solution 1
We draw the altitude to
, to get two right triangles.
![[asy] size(170); pair A = (0,0), B = (3, 0), C = (1, 4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B, 4)); label("$A$", A, S); label("$B$", B, S); label("$C$", C, N); label("$a$", (B+C)/2, NE); label("$b$", (A+C)/2, NW); label("$c$", (A+B)/2, S); label("$h$", (C+P)/2, E);[/asy]](http://latex.artofproblemsolving.com/c/a/7/ca7f7ca9366088b0274daace75e09ea38876ae3f.png)
Then , from the definition of the cotangent.
Let be the area of
Then
, so
.
By identical logic, we can find similar expressions for the sums of the other two cotangents:
Adding the last two equations, subtracting the first, and dividing by 2, we get
Therefore
Solution 2
By the law of cosines,
So, by the extended law of sines,
Identical logic works for the other two angles in the triangle. So, the cotangent of any angle in the triangle is directly proportional to the sum of the squares of the two adjacent sides, minus the square of the opposite side. Therefore
We can then finish as in solution 1.
Solution 3
We start as in solution 1, though we'll write instead of
for the area. Now we evaluate the numerator:
From the Law of Cosines and the sine area formula,
Then .
Solution 4
By the Law of Cosines,
Now
Solution 5
Use Law of cosines to give us or therefore
. Next, we are going to put all the sin's in term of
. We get
. Therefore, we get
.
Next, use Law of Cosines to give us . Therefore,
. Also,
. Hence,
.
Lastly, . Therefore, we get
.
Now, . After using
, we get
.
Solution 6
Let be
WLOG, assume that and
are legs of right triangle
with
and
By the Pythagorean theorem, we have , and the given
. Solving the equations gives us
and
. We see that
, and
.
Our derived equation equals as
approaches infinity.
Evaluating
, we get
.
Solution 7
As in Solution 1, drop an altitude to
. Let
meet
at
, and let
.
![[asy] size(170); pair A = (0,0), B = (3,0), C = (1,4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B , 4)); label("$A$", A, S); label("$B$", B, S); label("$C$", C, N); label("$P$", P, S); label("$x$", (A+P)/2, S); label("$y$", (B+P)/2, S); label("$a$", (B+C)/2, NE); label("$b$", (A+C)/2, NW); label("$c$", (A+B)/2, S); label("$h$", (C+P)/2, E);[/asy]](http://latex.artofproblemsolving.com/c/d/6/cd61091c8c8e5ee3e9ff7a2580f8c81786391de1.png)
Then, ,
. We can calculate
using the tangent addition formula, after noticing that
. So, we find that
\begin{align*}
\cot{\gamma} &= \frac{1}{\tan{\gamma}} \\
&= \frac{1}{\frac{\frac{x}{h} + \frac{y}{h}}{1 - \frac{xy}{h^2}}} \\
&= \frac{1}{\frac{(x+y)h}{h^2 - xy}} \\
&= \frac{h^2 - xy}{(x+y)h}.
\end{align*}
So now we can simplify our original expression: \begin{align*} \frac{\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac{\frac{h^2 - xy}{(x+y)h}}{\frac{x + y}{h}} \\ &= \frac{h^2 - xy}{(x+y)^2}. \end{align*}
But notice that , so this becomes
Now note that we can use the Pythagorean theorem to calculate
, we get that
So our expression simplifies to
since
from the problem and that there is another
after the
in our expression. Again note that
, so it again simplifies to
, or
.
Solution 8 (Quick and Easy)
Since no additional information is given, we can assume that triangle ABC is right with the right angle at B.
We can use the Pythagorean theorem to say
We can now solve for
in terms of
Using the definition of cotangent
Plugging into our desired expression, we get
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.