Official Solution:

\(x^2 + y^2 = 100\). All of the following could be true EXCEPT

A. \(|x| + |y| = 10\)
B. \(|x| \gt |y|\)
C. \(|x| \gt |y| + 10\)
D. \(|x| = |y|\)
E. \(|x| - |y| = 5\)


A. \(|x| + |y| = 10\) is possible if one is 0 and the other is 10.

B. \(|x| \gt |y|\) is possible if \(|x| \gt |5\sqrt{2}|\) and \(|y| \lt |5\sqrt{2}|\)

C. \(|x| \gt |y| + 10\) is never possible because if \(|x| \gt 10\), \(x^2+y^2\) becomes greater than 100, which is wrong.

D. \(|x| = |y|\) is possible if each is equal to \(|5\sqrt{2}|\).

E. \(|x| - |y| = 5\) is possible if \(|x| = |9.11|\) and \(|y| = |4.11|\).

Therefore all but C are possible. \(|x| \gt |y| + 10\) means \(x\) is greater than 10, which is not possible.


Answer: C
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Those are approximate values. Exact values are:
\(x = \frac{5}{2} (1+\sqrt{7})\), \(y = \frac{5}{2} (\sqrt{7}-1)\)
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Bunuel, can you please explain how you derive the exact values for x and y? Thanks.

30,000 Kudos. Awesome milestone.

Can this be solved visually??
1) as a circle on the xy plane
OR
2) as a tringle(pitagoras)

I think this the explanation isn't clear enough, please elaborate. please elaborate why C is right

and also why E is wrong?

for C : x^2 +y^2 -2xy>100 and we don't know anything about 2xy so how can we conclude C is right?

I thought of triangles and decided that C isn't possible although I think mapping the circle probably would have been quicker and less prone to error.

A, B and D can be quickly eliminated.

For A, we can have x=10, y=0. This makes B true as well immediately
For B, we can have x=y= 2root5

This leaves C and E. C looks easier to prove or disprove so start there. To maximise X, minimise Y. Abs val. Y cannot be negative so the smallest possible value for this is 0. That means the largest possible value for x is 10. X cannot be more than y+10 as y+10 = 10.

As we know this cannot be true, we don't even need to look at E