# Maths competition problems

1. Jul 19, 2008

### Kushal

1. The problem statement, all variables and given/known data

#1 If |x -2| < p, where x < 2, then x - p =

i have obtained an inequality at most.

#2 For what value of the coefficient a do the equations x2 -ax + 1 = 0 and x2 - x + a = 0.

#3 Let R be a rectangle. How many circles in the plane of R have a diameter both of whose endpoits are vertices of R?

3. The attempt at a solution

#1 -p < x-2 < p

eventually i obtain:

-2p + 2 < x - p < 2

#2 I have tried solving simultaneously but i have obtained and equation with both a and x.
i have also tried equating coefficients but i obtain a = 1. The actual answer is -2.

#3 is that 6? because there are 6 different ways of joining 2 vertices in the rectangle.

thnks

2. Jul 19, 2008

### HallsofIvy

That's all you can do. Certainly, given the conditions, x- p is not any specific number. For example, p=2, x= 1 and p= 2, x= 1/2 satisfy the conditions but in the first case x-p= -1 while in the second x-p= -3/2.

There is a verb missing! Those equation WHAT? Do you mean "for what value of a" are those equations true for the same x?

You haven't used the condition that x< 2. In fact, because of that x-2 is negative, |x-2|= 2- x and, since p is clearly positive, you should have -p< 2- x< p

What, exactly, is the question? You say the "actual answer" is -2. That would give x2- 2x= 1, which has 1 as its only root and x2- x+ 2 which has complex roots (1/2)+ i and (1/2)- i.

Yes, that sounds right. Any two points define a circle and there are 6 pairs of points. It is possible that in special cases two or more of those circles would be the same but the general case is 6.

thnks[/QUOTE]

Last edited by a moderator: Jul 19, 2008
3. Jul 20, 2008

### Kushal

sorry. the question is:

#2 For what value of the coefficient a do the equations x2 -ax + 1 = 0 and x2 - x + a = 0 have a common real solution?

for #3 the booklet says the answer is 5. i don't understand how. it is a rectangle. so the pair vertices are all different!

thanks!

4. Jul 20, 2008

### HallsofIvy

I see. One solution must be the same, not necessarily both. Try this. Suppose the solutions to the first equation are "u", and "v" while the solutions to the second equation are "u" and "w". Then you must have (x-u)(x-v)= x2- ax+ 1 and (x-u)(x-w)= x2- x+ a. Multiply those out and set like coeficients equal to get 4 equations for u, v, and w. What value of a gives a solution?

But it is a rectangle and not 4 arbitrary points. Take a look at the 2 circles having diagonally opposite points as ends of diameters.

5. Jul 28, 2008

### Kushal

thank you HallsofIvy, i got the solution to #2.

but i still cannot understand how come the two circles with the diagonals of the rectangle as diameter will be the same.

6. Jul 28, 2008

### HallsofIvy

The two diagonals of a rectangle have equal length and the same midpoint. A single circle with either diagonal as diameter goes through all four vertices and so has the other diagonal as diameter also.