Geometry + Distance

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by jaymw » Wed Dec 01, 2010 7:33 am
First one:

Calculate the runners' speeds first

Speed runner A: 24 miles/2 hours = 12 mph
Speed runner B: 24 miles/3 hours = 8 mph

Next, it totally doesn't matter in this problem whether the track is circular, straight, jagged, or a big looping (well okay that would be circular i guess).Anyway, it's enough to know distance and time to calculate a rate.

Now calculate when the runners meet: Treat the runners as one runner going the 24 miles at the speed of speed A + speed B. To cover 24 miles, that imaginary super runner will need 24miles/(12mph+8mph)=1.2 hours. That means the runners meet after 1.2 hours.

In 1.2 hours, runner A can runs 12mph*1.2hours=14.4 miles. That's the answer.

Second one:

The circumference of the circle is d*pi --> 12pi

15 degrees is 1/2 of the angle from the center to minor arc AC. That means minor arc AC equals 15*2/360*12pi=1pi

The same goes for minor arc BD because AB is parallel to CD and hence BCD is 15° as well. So, minor arc BD is 1 pi, too.

Minor arc AB is essentially the arc of the semicircle - minor arc AC - minor arc BD --> 6pi-1pi-1pi=4pi.

thus, d is right, but i guess you knew that already;)

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by N:Dure » Wed Dec 01, 2010 7:49 am
Good job explaining, but I don't get this part: "15 degrees is 1/2 of the angle from the center to minor arc AC." ?

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by jaymw » Wed Dec 01, 2010 8:14 am
An angle from the center to a minor arc is always twice the size of the angle from the intersection of the diameter and the circle to the same arc.

It's just a rule I learned and found to be helpful. I am not a mathematician, so I'm not able to derive it;)