Towns With Angles

Today's question also involves the law of cosines, but with a different purpose. It is a very neat question from a test which is considered "hard". Try it before looking at the answer.

A biker travels from town A to town B at 10 km/h. Another biker is traveling from town B to town C at 12 km/h. The distance from town A to town B is d. Angle ∠ABC = 120°. The bikers are closest after 2.5 hours of riding. Find d.

First, try to draw the situation. Draw the three towns, draw d, draw the angle, and place two points on segment AB and segment BC, which are the bikers.

Let t be the time the bikers are traveling. This means the first biker's distance from town A is 10t and the second biker's distance from town B is 12t. Since the first biker's distance from town A is 10t, his distance from town B is d - 10t.

We need a variable that represents the distance between the bikers. Let's call it x.

As you can see now, we have a triangle with three sides (x, 12t, d - 10t) and an angle of 120° which is opposite to the side with measure x. This situation calls for the law of cosines to find the relation of all variables:
x² = (12t)² + (d - 10t)² - 2(12t)(d - 10t)cos 120°

cos 120° = -0.5, so the relation becomes (after expanding):
x² = 144t² + d² - 20dt + 100t² + 12dt - 120t²

Add like terms and take the square root:
x = √(124t² + d² - 8dt)

We know that x has a minimum when t = 2.5, so if we derive x and plug 2.5 for t, we can find d.

As you should know, when you derive √f(x), you get f'(x)/2√f(x). Since we equate the derivative to zero (at a maximum or a minimum, the derivative is zero), we can now ignore the denominator (since it's always positive and we can multiply by it). Don't ignore it on a test, I'm doing it to save time.

Since we derive by t (dx/dt), the derivative of d² is 0 (d is a constant, not a variable).
0 = 248t - 8d

We know that t = 2.5, so:
0 = 620 - 8d
8d = 620
d = 77.5 km

Towns A and B are 77.5 km apart.

Have a great weekend,
Nadav

nadavs

Magic C

For today's question, I took another Yahoo Answers great question to give you the opportunity to solve a nice question. Here it comes:

The digits from 1 to 5 should be placed on a C shaped form like this:

x-x
x
x-x

(Each x represents a different digit)
The two horizontal rows (x-x) and the vertical row with three x's should have the same sum of digits.
In how many ways can this be done (reflections don't count as different), and why there are no other possibilities?

We can see at first that in order to fill in the digits, we need to find two pairs of digits that add up to the same number (the two x-x). There are three numbers that work like that: 5 (1+4, 2+3), 6 (2+4, 1+5), and 7 (3+4, 2+5).

However, 5 is impossible as a sum, since it leaves the digit 5 outside, which means it'll be in the middle x, creating a sum definitely greater than 5. We are left with only 6 and 7 as sums. After some playing with arrangements, we can see that the two solutions are:

For 6:
1-5
3
2-4

For 7:
2-5
1
4-3

There are no other possibilities because all other sums (except 5) have a maximum of one pair that works for them, and we already showed why 5 doesn't work.

Now it's your turn. Send in the riddles!

Have a great weekend,
Nadav

nadavs