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Difference between revisions of "1989 AIME Problems/Problem 10"

(→‎Solution)
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
−
We can draw the [[altitude]] <math>h</math> to <math>c</math>, to get two [[right triangle]]s. <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the [[cotangent]]. From the definition of area, <math>h=\frac{2A}{c}</math>, so <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}</math>.
+
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>
  
−
Now we evaluate the numerator:
+
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} = 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 ===
 +
 
 +
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>
 
<cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath>
Line 20: Line 63:
 
Then <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}</math>.
 
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 2 ===
+
=== Solution 4 ===
 
<cmath>\begin{align*}
 
<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}}
 
\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}}
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−
=== Solution 3===
+
=== 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>.
 
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>.
Line 45: Line 88:
  
  
−
=== Solution 4===
+
=== Solution 6===
  
  

Revision as of 20:43, 9 April 2022

Problem

Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find

$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$

Solution

Solution 1

We draw the altitude $h$ to $c$, 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]

Then $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$, from the definition of the cotangent.

Let $K$ be the area of $\triangle ABC.$ Then $h=\frac{2K}{c}$, so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}$.

By identical logic, we can find similar expressions for the sums of the other two cotangents: \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*} Adding the last two equations, subtracting the first, and dividing by 2, we get \[\cot \gamma = \frac{a^2 + b^2 - c^2}{4K}.\] Therefore \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} = 994. \end{align*}

Solution 2

By the law of cosines, \[\cos \gamma = \frac{a^2 + b^2 - c^2}{2ab}.\] So, by the extended law of sines, \[\cot \gamma = \frac{\cos \gamma}{\sin \gamma} = \frac{a^2 + b^2 - c^2}{2ab} \cdot \frac{2R}{c} = \frac{R}{abc} (a^2 + b^2 - c^2).\] 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 \[\frac{\cot \gamma}{\cot \alpha + \cot \beta} = \frac{a^2 + b^2 - c^2}{(-a^2 + b^2 + c^2) + (a^2 - b^2 + c^2)} = \frac{a^2 + b^2 - c^2}{2c^2}.\] We can then finish as in solution 1.

Solution 3

We start as in solution 1, though we'll write $A$ instead of $K$ for the area. Now we evaluate the numerator:

\[\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}\]

From the Law of Cosines and the sine area formula,

\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*}

Then $\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}$.

Solution 4

\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*}

By the Law of Cosines,

\[a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2\]

Now

\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*}


Solution 5

Use Law of cosines to give us $c^2=a^2+b^2-2ab\cos(\gamma)$ or therefore $\cos(\gamma)=\frac{994c^2}{ab}$. Next, we are going to put all the sin's in term of $\sin(a)$. We get $\sin(\gamma)=\frac{c\sin(a)}{a}$. Therefore, we get $\cot(\gamma)=\frac{994c}{b\sin a}$.

Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\cos(\beta)$. Therefore, $\cos(\beta)=\frac{a^2-994c^2}{ac}$. Also, $\sin(\beta)=\frac{b\sin(a)}{a}$. Hence, $\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}$.

Lastly, $\cos(\alpha)=\frac{b^2-994c^2}{bc}$. Therefore, we get $\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}$.

Now, $\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}$.


Solution 6

Let $\gamma$ be $(180-\alpha-\beta)$

$\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}$

WLOG, assume that $a$ and $c$ are legs of right triangle $abc$ with $\beta = 90^o$ and $c=1$

By Pythagorean theorem, we have $b^2=a^2+1$, and the given $a^2+b^2=1989$. Solving the equations gives us $a=\sqrt{994}$ and $b=\sqrt{995}$. We see that $\tan \beta = \infty$, and $\tan \alpha = \sqrt{994}$.

We see that our derived equation equals to $\tan^2 \alpha$ as $\tan \beta$ approaches infinity. Evaluating $\tan^2 \alpha$, we get $\boxed{994}$.

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

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