1989 AHSME Problems/Problem 13

Revision as of 06:51, 22 October 2014 by Timneh (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Two strips of width 1 overlap at an angle of $\alpha$ as shown. The area of the overlap (shown shaded) is

[asy] pair a = (0,0),b= (6,0),c=(0,1),d=(6,1); transform t = rotate(-45,(3,.5)); pair e = t*a,f=t*b,g=t*c,h=t*d; pair i = intersectionpoint(a--b,e--f),j=intersectionpoint(a--b,g--h),k=intersectionpoint(c--d,e--f),l=intersectionpoint(c--d,g--h); draw(a--b^^c--d^^e--f^^g--h); filldraw(i--j--l--k--cycle,blue); label("$\alpha$",i+(-.5,.2)); //commented out labeling because it doesn't look right. //path lbl1 = (a+(.5,.2))--(c+(.5,-.2)); //draw(lbl1); //label("$1$",lbl1);[/asy]

$\textrm{(A)}\ \sin\alpha\qquad\textrm{(B)}\ \frac{1}{\sin\alpha}\qquad\textrm{(C)}\ \frac{1}{1-\cos\alpha}\qquad\textrm{(D)}\ \frac{1}{\sin^{2}\alpha}\qquad\textrm{(E)}\ \frac{1}{(1-\cos\alpha)^{2}}$

Solution

[asy] pair a = (0,0),b= (6,0),c=(0,1),d=(6,1); transform t = rotate(-45,(3,.5)); pair e = t*a,f=t*b,g=t*c,h=t*d; pair i = intersectionpoint(a--b,e--f),j=intersectionpoint(a--b,g--h),k=intersectionpoint(c--d,e--f),l=intersectionpoint(c--d,g--h); draw(a--b^^c--d^^e--f^^g--h); filldraw(i--j--l--k--cycle,blue); label("$\alpha$",i+(-.4,.15),fontsize(8)); label("$\alpha$",i+(.4,-.15),fontsize(8)); draw(j--t*j); draw(rightanglemark(j,t*j,i), linewidth(0.5)); path lbl1 = (a+(1.5,.05))--(c+(1.5,-.05)); draw(lbl1,Arrows); label("$1$",lbl1);[/asy]

The rhombus has a base of length $\frac1{\sin\alpha}$ and height of $1$. Its area is the product.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS