How Do You Solve These Challenging Math Competition Problems?

[Kushal]

Homework Statement

#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?

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

 

[HallsofIvy]

Homework Statement

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

i have obtained an inequality at most.p

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.

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

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

#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?

The Attempt at a Solution

#1 -p < x-2 < p

eventually i obtain:

-2p + 2 < x - p < 2

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

#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.

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.

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

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]

 

[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!

 

[HallsofIvy]

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?

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?

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!

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.

 

[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.

 

[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.