2010 Galois Contest

I took the Galois contest this morning, it was a 75 minute test, with calculators allowed. It has four questions, each worth ten points, and a full written response is required.

I’m going to post my answers here. These are my own answers, so no guarantee that they’re correct.

Although some schools haven’t took it yet, it should still be fine. Nobody would care if I leak answers. Yay me. Anyways, first question.

Problem 1

1. Emily has an old showerhead using 18L of water per minute and a new showerhead using 13L a minute.

a) How long does it take to use 260L of water using the new showerhead?

13L per minute and 260L of water, so 20 minutes.

b) How much less water is used in 10 minutes with the new showerhead than the old one?

In 10 minutes the old showerhead uses 180L of water and the new one only 130L.

So the new one uses 50L less water.

c) Water costs 8 cents per 100L. How much money is saved by using the new showerhead instead of the old one, in a 15 minute shower?

In 15 minutes the old showerhead uses 270L of water and the new one uses 195. So the difference is 75L which costs 6 cents.

d) How long (in minutes) does it take to save $30 by using the new showerhead?

If m is the time showered in minutes then the cost of the old shower per minute (in cents) is or and the new shower or , and so the difference is .

If we save $30 then we save 3000 cents, and and m = 7500 minutes.

In the contest I somehow thought $30 was 30000 cents and not 3000, so I was off by an order of magnitude. I wonder how much I’ll still get for that question.

Problem 2

2 a) Determine the area of this quadrilateral:

The area of a trapezoid is where and are the two parallel sides.

Using this formula we get or 84.

b)

Determine an expression for the area of in terms of p.

The area of a triangle is , and by substitution the area of COB is or 6p.

From now on I will use the notation [ABC] to describe the area of triangle ABC.

c) Determine an expression for [QCA] in terms of p.

Using QC as the height, QC can be represented by . Substituting into the area formula for a triangle again, the expression should be 12-p.

d) If [ABC], determine the value of p.

Use what we got before. Notice that [ABC] = [QABO] – [COB] – [QCA], which we determined to be or .

We know that so we solve the equation for p, and get p=9.

Problem 3

3 a) Solve this system for :

Using elimination or substitution, you should get (x,y) = (26,16).

b) If p is an even integer and q is an odd integer, prove that the following system of equations,

has no positive integer solutions.

By subtraction, we can get . The left hand side is even so the right hand side must be even. But clearly the right hand side is odd (even integer minus odd integer). Contradiction.

c) Determine all solutions to the equation for (x,y) being positive integers.

We can rewrite the equation as , and let p=x+y and q=x-y. From what we’ve seen in (b), p-q has to be even so p and q must be both even, or both odd. Because pq=420, p and q are both even (odd * odd != even).

There are four pairs of even integer pairs (p,q) satisfying pq=420: (2,210), (6,70), (10,42), and (14,30).

The four pairs each give the solutions (104,102), (38,32), (26,16), and (22,8) for (x,y).

Problem 4

4 a)

Explain why [PQT]:[PTR] = 3:5.

The two triangles have different bases but share the same altitude, or height. Thus the ratio of their areas is or 3:5.

b)

Here BD=DC; AE:EC=1:2; AF:FD=3:1. If [DEF]=17 determine [ABC].

Following a similar logic as in part (a), we know that [ABD]=[ADC], [DAE]:[DEC]=1:2, and [EAF]:[EFD]=3:1.

Since [EFD]=17, [EAF]+[EFD]=4*17 or 68. So [DAE]=68, and [DAE]+[DEC]=3*68 or 204. Finally [ADC]=204, so [ABC]=2*204 = 408.

This is just the exact same logic repeated three times.

c)

Here, VY:YW = 4:3, [PYW]=30, and [PZW]=35.

Because VY:YW = 4:3, we can find [VPY] to be 40 the same way that we did it in part (b).

In , and , so the ratio is 2:1. Therefore the ratio VP:PZ = 2:1.

In order to simplify things a little bit, let , , and .

Here’s a diagram, with new labels:

In the whole triangle, [UVY]:[UYW] = 4:3. We can write this algebraically:

Now using the line XW as a base, the ratios of the triangles on either side is equal. Namely, [XVP]:[PVW] = [UXP]:[UPW]. Writing this algebraically:

Finally we use VZ as a base. We know the ratio to be 2:1, so writing it algebraically would be this:

Solving the three equations for the three variables, we get (x,y,z) = (56,84,70). [UXP]=y, so [UXP]=84.

I was unable to get this far on the contest, and this problem is by far the hardest in the set. Looking back, I was actually on the right track, but gave up my approach thinking it was completely wrong.

If I’m lucky, I might get some part marks for my work here.

I think I did better on this contest than I did on the Cayley. Probably still won’t win any prizes, though.

In other news

I’ve placed in the first group in the Canadian Computing Competition, along with Wen Li Looi from my school.