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

m (See also)
(Solution 8)
 
(15 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
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=1989^{}_{}c^2</math>, find
+
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 can draw the altitude h to c, to get two right triangles.
+
=== 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]].
  
From the definition of area, <math>h=\frac{2A}{c}</math>, so therefore <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}</math>
+
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>.
  
Now we evaluate the numerator:
+
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>
  
<math>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</math>.
+
=== Solution 2 ===
  
<math>\cos{\gamma}=\frac{1988c^2}{2ab}</math>, from the [[Law of Cosines]]
+
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.
  
<math>\sin{\gamma}=\frac{c}{2R}</math>, where R is the circumradius.
+
=== Solution 3 ===
  
<math>\cot{\gamma}=\frac{1988cR}{ab}</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:
  
Since <math>R=\frac{abc}{4A}</math>, <math>\cot{\gamma}=\frac{1988c^2}{4A}</math>
+
<cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath>
  
<math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=994</math>
+
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>cotγcotα+cotβ=cotγsinαsinβsinγ=cosγsinαsinβsin2γ=abc2cosγ=abc2994c2ab=994</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}}
 
{{AIME box|year=1989|num-b=9|num-a=11}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
[[Category:Intermediate Trigonometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:41, 20 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 cotγ=1tanγ=1xh+yh1xyh2=1(x+y)hh2xy=h2xy(x+y)h.

So now we can simplify our original expression: cotγcotα+cotβ=h2xy(x+y)hx+yh=h2xy(x+y)2.

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


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 \[c^2+a^2=b^2\] We can now solve for $a$ in terms of $c$

\[c^2+a^2=1989c^2-a^2\] \[a^2=994c^2\] \[a=\sqrt{994}c\]

Using the definition of cotangent

\[cot(A)=\frac{c}{a}=\frac{1}{\sqrt{994}}\] \[cot(B)=cot(90)=0\] \[cot(C)=\frac{a}{c}=\sqrt{994}\] Plugging into our desired expression, we get $\boxed{994}$

~ms0001

See also

1989 AIME (ProblemsAnswer KeyResources)
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. AMC logo.png