In triangle ABC, ∠A = 90°. Point X on BC is such that ∠XAB = 45° and CX = BX - AX = 1. Find AC and AX?

Good suggestion JJWJ, that got ME to the solution. However, you might include a mention of ∠XAB = 45°, or (x-coordinate of "X" = y-coordinate of "X"), in your hints. That equation together with CX = BX - AX = 1 makes a (nasty) solvable system. Too nasty. Once I had AX, I found the minimal polynomial for that value (over the integers) and went looking for THAT equation in the picture. Here is my ?better? solution:

Consider the perpendicular from X to AB, meeting AB at D, and the perpendicular from X to AC, meeting AC at E. Then right triangles XDB and CEX are similar (both similar to CAB) and, since ∠XAB = 45°, DX = EX = AX/√2. Also CX=1, BX=1+AX and, by Pythagoras, BD=√(AX²/2+2AX+1). Hence, CX/EX=XB/DB, producing the equation:

1 / (AX/√2) = (1+AX) / (√(AX²/2 + 2AX + 1) or (AX)^4 + 2(AX)^3 - 4(AX) - 2 = 0

Which produces (with some effort!) a unique positive root:

AX = [FourthRoot(12) + SquareRoot(3) - 1] / 2

Given that this (easily obtained polynomial) is the minimal polynomial for AX over the integers, I couldn't imagine anything much simpler, until I calculated AC.

I'm getting AC = FourthRoot(3).

So the equation [AC^4-3=0] should be hiding somewhere in there?!?

I haven't found it yet ... Anyone?

(I do not know what "surd form" means.)

I am assuming that you are writing that the angle BAC is equal to ninety degrees.

If this is correct, then place the triangle in the first quadrant with A = (0, 0), B = (0, b) and C = (c, 0) where b and c are positive.

1) Determine the equation of the line in the hypotenuse of the right triangle

(connecting B and C).

2) Write down "X = ( , )" filling in the two coordinates with variables. Since X is on that one line, then you will not use two different variables (like a and b) when you fill in the coordinates, but use only one variable.

3) Given 1) and 2) above, find the fomula for CX, BX and AX.

4) Now, solve your problem.

That's cool! Happy Belated Birthday to Late Voldemort! Happy New Year to yeh all!