How long is the belt? - May 20-24, 1996
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I was playing in a field hockey tournament this weekend, hoping a great situation for the problem of the week would magically appear, but nothing happened until I was pulling out of the parking lot and my car made an awful noise it's been making every once in a while. I know I have to get the belts replaced, so we're going to figure out how long the belt is. (I could just look it up in the parts catalog, but that wouldn't make a very interesting problem.)There are two wheels. The belt runs around one and crosses over before going around the other one, so the wheels will turn in opposite directions. (This is probably so people like me will get confused and pay someone else a lot of money to fix their cars!)
If you follow the path of the belt, it starts at the top of the big wheel, goes down to the bottom of the small wheel, up and around, and then down to the bottom of the big wheel and back around to the top.
The diameter of the big wheel is 18 cm, and the diameter of the small one is 12 cm. The centers of the two wheels are 30 cm apart, and the belt crosses 18 cm from the center of the big wheel.
How long is the belt? How do you know you're right?
Annie says: Solutions
What a week: 45 right answers and 57 wrong, and 23 people got the exact same wrong answer! This caused me to reevaluate my own effort, but then I figured out what was going on, and it's a bummer that so many folks were making the same wrong assumptions, which are so clearly impossible. Yes, I realize they might not have been "clearly impossible" to everyone, but some of the folks who got it wrong had pictures that showed that it wouldn't work!
Here's what those 23 people did wrong. They assumed that the belts were only in contact with half of the wheels. Why isn't this true? Because we know that the belts are tangent to the wheels at some point, and that anything tangent to a circle is perpendicular to the radius at that point. So if we have two pieces of the belt coming off the wheel halfway around, then both pieces are perpendicular to the same diameter, and so the two belts are parallel, and they'll never cross! And if you draw them so that they do cross, right from the tops and bottoms of the wheels, then the belts cut through the wheels, which just won't work - this is what a number of folks drew, and they still figured it would work.
So that lost them some length - which they almost gained back when figuring the lengths of the straight parts of the belts. What initially confused me about the answers was that while the right answer is 114.79 cm, the common wrong answer was 114.206 cm. Those answers are pretty close.
If you draw a correct picture (with the belts covering about 2/3 of the wheels), you see that we still have right triangles, but the right angle is at the point of tangency, and since the radius is 9 and the hypotenuse is 18, we have a 30-60-90 triangle, and so everything is very neat and tidy - you don't even need the Pythagorean theorem if you remember enough about 30-60-90 triangles.
Is there a moral to all of this? I guess it's just to think things through carefully and see if your assumptions and your pictures make sense before doing all of the math.
Following are some highlighted solutions. The names of all the people who submitted correct solutions and most of the solutions are also available. There are some very nice ones, so I encourage you to read through them. We also have two new schools, including our first solution from Italy!
Thomas S. Kuo Grade: 7 School: Murray Junior High School, Ridgecrest, California A * * * B * * C * * * * * * * D * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * E O F E is the center of the big wheel with diameter 18 cm. F is the center of the small wheel with diameter 12 cm. O is the cross point of line between the top of the big wheel to the bottom of the small wheel and line between the top of the small wheel and the bottom of the big wheel. Line OB is tangent to the big wheel at point B. Line OC is tangent to the small wheel at point D. angle AEO = angle EBO = angle OFC = angle FDO = 90 degrees EF = 30 cm EO = 18 cm, EA = EB = 9 cm FO = 12 cm, FC = FD = 6 cm For the big wheel: 1. EB^2 + BO^2 = EO^2 9^2 + BO^2 = 18^2 BO = sqrt(243) ~ 15.59 cm 2. The distance around the semicircle of the big wheel is (1/2) * 18 pi = 9 pi 3. angle AEB = angle BOE = arc sin (BE/EO) = arc sin (9/18) = arc sin (1/2) = 30 degrees arc AB = 18 pi * (30/360) = (3/2) pi cm 4. The total length starts O, goes around the big wheel, returns to O is 2 * (15.59 + (3/2) pi) + 9 pi = 31.18 + 12 pi ~ 68.88 cm Similarly for the small wheel: 5. OD^2 + DF^2 = OF^2 OD^2 + 6^2 = 12^2 OD = sqrt(108) ~ 10.39 cm 6. The distance around the semicircle of the small wheel is (1/2) * 12 pi = 6 pi 7. angle CFD = angle DOF = arc sin (DF/OF) = arc sin (6/12) = arc sin (1/2) = 30 degrees arc CD = 12 pi * (30/360) = pi 8. The total length starts O, goes around the small wheel, returns to O is 2 * (10.39 + pi) + 6 pi = 20.78 + 8 pi ~ 45.92 cm Length of the belt: 9. 68.88 + 45.92 ~ 114.80 cm
Greg Paul Grade 12 Fairfield High School, Fairfield, CT The length of the belt is 114.793 centimeters. First draw a line from the center of the circle with radius 9 to its tangent point. We know that the angle between that line and the line from the tangent to where the belts cross is 90 degrees. Let the center of the circle be point A, the tangent point be B and where the belt crosses be C. We know AB is 9 and AC is 18, then we know that inverse sin (9/18) is 60 degrees; therefore, triangle ABC is a 30-60-90 triangle. Then we can set up a proportion to find the length of the belt around the wheel. 360/18pi = 60/x. x turns out to be 3pi. Double 3pi and get 6pi. This is the length where the belt does not touch the wheel. Subtract this from 18pi giving you 12pi. The length of the hypotenuse of triangle ABC is 9*sqrt(3) because it is the radius times sqrt(3) in a 30-60-90 triangle. Double that to get 18*sqrt(3) for both sides of the belt since tangent segments from the same external point are equal. Set up proportions to get the measurements for the other circle. Total length of the belt is 12pi + 8pi + 18*sqrt(3) + 12*sqrt(3) = 114.793 centimeters.
Rita Beckner Grade 9 Franklin County High School, Rocky Mount, Virginia It sounds as though your car really has a problem. It would be much easier to look in a parts manual to find the length of the belt, but if you don't have one, here is how to find the length of the belt: Statements Reasons NB=18, OK=12 given CK=6, AB=9 diameter = 2radius JA=9, CI=6 radii of a circle are congruent JK is tangent to c1, c2 given BI is tangent to c1, c2 given AB is perp. to BI, AJ is perp. to JK radii perp. to intersecting tangents JK is perp. to KC, CI is perp. to IB radii perp. to intersecting tangents angle B, angle J, angle K, angle I rt. perp lines form rt. angles Measure angle B=90, measure angle J=90 rt. angles are 90 degrees measure angle K=90, measure angle I=90 rt. angles are 90 degrees AE=18 given AC=AE+EC segment addition postulate AC=30 given 30=18+EC substitution 12=EC subtraction of equality EK squared+KC squared=EC squared Pythagorean theorem CI squared+IE squared=CE squared Pythagorean theorem AB squared+BE squared=AE squared Pythagorean theorem AJ squared+JE squared=AE squared Pythagorean theorem EK squared+6 squared=12 squared substitution IE squared+6 squared=12 squared substitution BE squared+9 squared=18 squared substitution JE squared+9 squared=18 squared substitution EK squared+36=144 substitution 36+EK squared=144 substitution BE squared+81=324 substitution JE squared+81=324 substitution JE+9 radical 3 substitution BE=9 radical 3 substitution IE=6 radical 3 substitution EK=6 radical 3 substitution cosine BAM, JAM= 9/18 cosine=adjacent leg/hypotenuse measure angle BAM, measure angle JAM=60 Definition cosine cosine LCK, LCI=6/12 cosine=adjacent leg/hypotenuse measure angle BAM+ JAM=measure angle BAJ angle addition postulate measure angle LCK+ LCI=measure angle KCI angle addition postulate 60+60=measure angle BAJ substitution 60+60=measure angle KCI substitution measure angle BAJ, KCI = 120 substitution circle C, A = 360 degrees degrees of a circle = 360 360-120=arc BPJ measure of major arc=360-minor arc 360-120=arc KDI measure of major arc=360-minor arc arc BPJ, KDI=240 substitution 240/360= portion of circle=major arc substitution pi(18)(2/3)= major arc perimeter circumference=pi(diameter) pi(12)(2/3)= major arc perimeter circumference=pi(diameter) 37.7, 25.5=arc perimeter substitution arc BPJ+arc KDI+IE+EK+EB+JE= segment addition postulate length of belt 37.7+25.1+6 radical 3+6 radical 3+ substitution 9 radical 3+9 radical 3= length of belt length of belt=114.8 centimeters substitution
From: peterson@eng16.rochny.uspra.abb.com Dave Peterson Grade 5 homeschool teacher Rochester, NY It turns out that we were given one extra piece of information in this problem, namely the location of the belt crossing, which isn't really needed and which I can figure out from the rest of the data. The car manufacturer also was nice enough to give us a very convenient shape, so we don't have to use any inverse trigonometric functions as I would have expected. I'll start with the straight parts of the belt. If I draw just one of these, I see two (similar) triangles bounded by the radii of the wheels, the line between the centers, and the belt (which is tangent to both circles, and therefore perpendicular to both radii): /\ /\/ \ 9 / \ belt / \ 12 /________________\____________ 18 \ / \ /\/ 6 \/ But rather than work with these two triangles, I can change it into just one big triangle by constructing a rectangle above the diagonal belt: /\ 6 /\/ \ / \ /\ \ =belt /\/ \ \ 9 / \ belt \ / \ \ /________________\___________\ 30 \ / \ /\/ 6 \/ Now I don't need to use the distance to the crossing, but only the two radii, which give 15 cm for the length of one leg, and the distance between the wheels, which gives 30 cm for the hypotenuse. The length of each straight section is then sqrt(30^2 - 15^2) = sqrt(675) = sqrt(15*15*3) = 15 * sqrt(3) Next, I notice that with the hypotenuse twice as long as the leg and this triangle is half of an equilateral triangle, so the angle at left is 60 degrees. The arc of each wheel taken up by the belt is therefore 360 - 2 * 60 = 240 degrees, and the length of each arc is 2/3 of the circumference of the wheel (2*PI*r), or 2*PI*9*2/3 and 2*PI*6*2/3 for the left and right wheel respectively. The total length therefore is the sum of two equal straight sections and two different arcs: 30*sqrt(3) + 12*PI + 8*PI = 30*1.7321 + 20*3.1416 = 114.79 cm (As a sanity check, the perimeter of a rectangle around the whole belt would be 18 by 45, or 126 cm, which is close.) For extra credit, to determine the location of the crossing if it were not given (or to check the value that was given), the two similar triangles in my figure give (with x = distance from the center of the left wheel to the crossing) x 30 - x --- = ------ 9 6 so 6 * x = 270 - 9 * x 15 * x = 270 x = 18 cm as given.
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