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

(→‎Solution 1)
(Added Solution 7)
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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>
 
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 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>.  
+
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>.  
  
−
We see that our derived equation equals to <math>\tan^2 \alpha</math> as <math>\tan \beta</math> approaches infinity.
+
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>.
 
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]]
  
 
== See also ==
 
== See also ==

Revision as of 15:45, 18 January 2024

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} = \boxed{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 the 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 $\tan \alpha = \sqrt{994}$.

Our derived equation equals $\tan^2 \alpha$ as $\tan \beta$ approaches infinity. Evaluating $\tan^2 \alpha$, we get $\boxed{994}$.


Solution 7

As in Solution 1, drop an altitude $h$ to $c$. Let $h$ meet $c$ at $P$, and let $AP = x, BP = y$.

[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]

Then, $\cot{\alpha} = \frac{1}{\tan{\alpha}} = \frac{x}{h}$, $\cot{\beta} = \frac{1}{\tan{\beta}} = \frac{y}{h}$. We can calculate $\cot{\gamma}$ using the tangent addition formula, after noticing that $\cot{\gamma} = \frac{1}{\tan{\gamma}}$. 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 $x+y = c$, so this becomes \[\frac{h^2 - xy}{c^2}.\] Now note that we can use the Pythagorean theorem to calculate $h^2$, we get that \[h^2 = \frac{a^2 - y^2 + b^2 - x^2}{2}.\] So our expression simplifies to \[\frac{1989c^2 - (x+y)^2}{2c^2}\] since $a^2 + b^2 = 1989c^2$ from the problem and that there is another $-\frac{2xy}{2}$ after the $h^2$ in our expression. Again note that $x+y = c$, so it again simplifies to $\frac{1988c^2}{2c^2}$, or $\boxed{994}$.

~Yiyj1

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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