Difference between revisions of "1989 AHSME Problems/Problem 13"
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| − | + | == Problem == | |
| − | <math> \ | + | Two strips of width 1 overlap at an angle of <math>\alpha</math> 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> | ||
| + | |||
| + | <math> \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}} </math> | ||
| + | |||
| + | == 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 <math>\frac1{\sin\alpha}</math> and height of <math>1</math>. Its area is the product. | ||
Revision as of 08:46, 24 September 2012
Problem
Two strips of width 1 overlap at an angle of 
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]](http://latex.artofproblemsolving.com/7/6/2/762566bf70020b367302b9e38d58b5e686d1c83c.png)
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]](http://latex.artofproblemsolving.com/b/9/2/b92b7b2728ab336aa519c33818d4f502540198b6.png)
The rhombus has a base of length 
and height of 
. Its area is the product.
