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Difference between revisions of "1992 USAMO Problems/Problem 2"
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<math>\frac {\sin{1}}{\cos{0}\cos{1}} + \frac {\sin{1}}{\cos{1}\cos{2}} + ... + \frac {\sin{1}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math>
 
<math>\frac {\sin{1}}{\cos{0}\cos{1}} + \frac {\sin{1}}{\cos{1}\cos{2}} + ... + \frac {\sin{1}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math>
  
−
we can right this as:
+
we can write this as:
  
 
<math>\frac {\sin{1 - 0}}{\cos{0}\cos{1}} + \frac {\sin{2 - 1}}{\cos{1}\cos{2}} + ... + \frac {\sin{89 - 88}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math>
 
<math>\frac {\sin{1 - 0}}{\cos{0}\cos{1}} + \frac {\sin{2 - 1}}{\cos{1}\cos{2}} + ... + \frac {\sin{89 - 88}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math>

Revision as of 11:35, 8 August 2010

Problem

Prove \[\frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}.\]

Solution 1

Consider the points $M_k = (1, \tan k^\circ)$ in the coordinate plane with origin $O=(0,0)$, for integers $0 \le k \le 89$.

[asy] size(200); defaultpen(1); pair O=(0,0), a=expi(0), b=expi(1/6), c=expi(2/6), d=expi(3/6), y=expi(32/30), z= expi(34/30); pair A=a, B=b/b.x, C= c/c.x, D=d/d.x, Y=y/y.x, Z=z/z.x, E=(D+Y)/2; pair X=(O+E)/2; draw(O--A--Z); draw(B--O--C--D--O--Y--Z--O); label("\(O\)",O,SW); label("\(M_0\)",A,ESE); label("\(M_1\)",B,ESE); label("\(M_2\)",C,ESE); label("\(M_3\)",D,ESE); label("\(\vdots\)",E,ESE); label("\(M_{88}\)",Y,ESE); label("\(M_{89}\)",Z,ESE); label("\(\ddots\)",X); [/asy]

Evidently, the angle between segments $OM_a$ and $OM_b$ is $(b-a)^\circ$, and the length of segment $OM_a$ is $1/\cos a^\circ$. It then follows that the area of triangle $M_aOM_b$ is $\tfrac{1}{2} \sin(b-a)^\circ \cdot OM_a \cdot OM_b = \tfrac{1}{2} \sin(b-a)^\circ / (\cos a^\circ \cdot \cos b^\circ)$. Therefore \begin{align*} \sum_{k=0}^{88} \frac{\tfrac{1}{2} \sin 1^\circ}{ \cos k^\circ \cos k+1^\circ} &= \sum_{k=0}^{88} [ M_k O M_{k+1} ] \\ &= [ M_0 O M_{89} ] \\ &= \frac{ \tfrac{1}{2} \sin 89^\circ}{\cos 89^\circ} = \frac{\tfrac{1}{2} \cos 1^\circ}{\sin 1^\circ} , \end{align*} so \[\sum_{k=0}^{88} \frac{1}{\cos k^\circ \cos (k+1)^\circ} = \frac{\cos 1^\circ}{ \sin^2 1^\circ},\] as desired. $\blacksquare$

Solution 2

First multiply both sides of the equation by $\sin 1$, so the right hand side is $\frac{\cos 1}{\sin 1}$. Now by rewriting $\sin 1=\sin((k+1)-k)=\sin(k+1)\cos(k)+\sin(k)\cos(k+1)$, we can derive the identity $\tan(n+1)-\tan(n)=\frac{\sin 1}{\cos(n)\cos(n+1)}$. Then the left hand side of the equation simplifies to $\tan 89-\tan 0=\tan 89=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}$ as desired.

Solution 3

Multiply by $\sin{1}$. We get:

$\frac {\sin{1}}{\cos{0}\cos{1}} + \frac {\sin{1}}{\cos{1}\cos{2}} + ... + \frac {\sin{1}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}$

we can write this as:

$\frac {\sin{1 - 0}}{\cos{0}\cos{1}} + \frac {\sin{2 - 1}}{\cos{1}\cos{2}} + ... + \frac {\sin{89 - 88}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}$

This is an identity $\tan{a} - \tan{b} = \frac {\sin{(a - b)}}{\cos{a}\cos{b}}$

Therefore;

$\sum_{i = 1}^{89}[\tan{k} - \tan{(k - 1)}] = \tan{89} - \tan{0} = \cot{1}$, because of telescoping.

but since we multiplied $\sin{1}$ in the beginning, we need to divide by $\sin{1}$. So we get that:

$\frac {1}{\cos{0}\cos{1}} + \frac {1}{\cos{1}\cos{2}} + \frac {1}{\cos{2}\cos{3}} + .... + \frac {1}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin^2{1}}$ as desired. QED

Resources

1992 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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