BuilderBoi

I think it is possible...

Let \(\angle AXE = \angle CYX = x\)

We have the equation: \(x = 3x-98\)

Solving for x, we find that \(x = 49\).

Now we have: \( \angle AXE = \angle BXY = \color{brown}\boxed{49}\)

What am I missing?

Thanks for answering everyone!!

Nor do I!

Because the number must be a multiple of 5, the final digit must be a 5. 

There are \(5! = 120\) ways to order the other digits. 

For the number to \(>450,000\), the numbers can be in the order \(46\text{_,__}5\) or \(6 \text{__,__5}\)

For the first case, there are \(3! = 6\) ways to order the 3 remaining digits. 

For the second case, there are \(4! = 24\) ways to order the remaining 4 digits. 

Thus, the probability is \({24 + 6 \over 120} = \color{brown}\boxed{1 \over 4}\)

No, but I found the answer already by guessing and checking, \(x = 16\), \(y = 15\) and \(z = 3\)

That's interesting, I wonder why it works....

I did that, and I just brute forced it from here...

Surely there's a better way...

Problem 1) PQ is similar to AB because of Side-Angle-Side. P and Q lie on sides AO and OB respectively, and \(\angle POQ = \angle AOB\).

Problem 2) Because \(OP = {3 \over 5} \times AO\), we know that \(OP = {3 \over 5} \times 10 = \color{brown}\boxed{6}\)

Because \(PN = 8 \), \(QN = 10\), and \(\angle N = 90\), \(PQ = \sqrt{164} = 2\sqrt{41}\)

We know that \(M\) is the midpoint, so \(PM = \sqrt{41}\).

Now, draw \(MA\), so that \(MA \) is perpendicular to \(PR\).

Because of similar triangles, \(MA = 5\) and \(AN = 4\)

Using the Pythagorean Theorem, we find that \(RM = 13\).

Because of similar triangles, \(OR = {2 \over 3} \times 13 = \color{brown}\boxed{26\over3}\) 

Wait... I made an error, I wrongfully concluded that PF was 1/3 BN. MaxWong's answer is correct, and mine is incorrect