1992 USAMO Problems/Problem 2

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

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. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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

Discussion on AoPS/MathLinks

Page

Toolbox

Search

1992 USAMO Problems/Problem 2

Problem

Solution

Resources