Difference between revisions of "1989 AHSME Problems/Problem 13"

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Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers with <math>a<2b</math>, <math>b<3c</math>, and <math>c<4d</math>. If <math>d<100</math>, the largest possible value for <math>a</math> is
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== Problem ==
  
<math> \mathrm{(A) \ 2367 } \qquad \mathrm{(B) \ 2375 } \qquad \mathrm{(C) \ 2391 } \qquad \mathrm{(D) \ 2399 } \qquad \mathrm{(E) \ 2400 } </math>
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Two strips of width 1 overlap at an angle of <math>\alpha</math> as shown. The area of the overlap (shown shaded) is
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<asy>
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pair a = (0,0),b= (6,0),c=(0,1),d=(6,1);
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transform t = rotate(-45,(3,.5));
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pair e = t*a,f=t*b,g=t*c,h=t*d;
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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);
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draw(a--b^^c--d^^e--f^^g--h);
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filldraw(i--j--l--k--cycle,blue);
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label("$\alpha$",i+(-.5,.2));
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//commented out labeling because it doesn't look right.
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//path lbl1 = (a+(.5,.2))--(c+(.5,-.2));
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//draw(lbl1);
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//label("$1$",lbl1);</asy>
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<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>
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== Solution ==
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<asy>
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pair a = (0,0),b= (6,0),c=(0,1),d=(6,1);
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transform t = rotate(-45,(3,.5));
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pair e = t*a,f=t*b,g=t*c,h=t*d;
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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);
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draw(a--b^^c--d^^e--f^^g--h);
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filldraw(i--j--l--k--cycle,blue);
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label("$\alpha$",i+(-.4,.15),fontsize(8));
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label("$\alpha$",i+(.4,-.15),fontsize(8));
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draw(j--t*j);
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draw(rightanglemark(j,t*j,i), linewidth(0.5));
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path lbl1 = (a+(1.5,.05))--(c+(1.5,-.05));
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draw(lbl1,Arrows);
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label("$1$",lbl1);</asy>
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The rhombus has a base of length <math>\frac1{\sin\alpha}</math> and height of <math>1</math>. Its area is the product.
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== See also ==
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{{AHSME box|year=1989|num-b=12|num-a=14}} 
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[[Category: Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 07:51, 22 October 2014

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

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