24th Eötvös Competition Problems 1917

1.  a, b are integers. The solutions of y - 2x = a, y² - xy + x² = b are rational. Show that they must be integers.

2.  A square has 10s digit 7. What is its units digit?

3.  A, B are two points inside a given circle C. Show that there are infinitely many circles through A, B which lie entirely inside C. 

Solutions

Problem 1

a, b are integers. The solutions of y - 2x = a, y² - xy + x² = b are rational. Show that they must be integers.

Solution

We have x = (y-a)/2, substituting in the other equation gives 3y² = 4b-a². Hence (3y)² = 3(4b-a²), which is an integer, so 3y is an integer. Hence 3(4b-a²) is a square, so (4b-a²) must be divisible by 3. Hence y² = (4b-a²)/3 is an integer, so y is an integer.

Since 3y² + a² = 4b, which is even, y and a must have the same parity. Hence y-a is an even integer, and so x is an integer.

Note that substituting y = a+2x does not work so easily.

Problem 2

A square has 10s digit 7. What is its units digit?

Solution

Suppose n = 100m + 10a + b. Then n² = 20ab + b² mod 100. So we are only interested in the last two digits of n. Also we require that b2 has an odd carry. Checking 0, 1, 4, 9, 16, 25, 36, 49, 64, 81 we see that only 16 and 36 have an odd carry, and in both cases they give units digit 6. So the units digit must be 6. Note that there are such squares, eg 24² = 576. 

Problem 3

A, B are two points inside a given circle C. Show that there are infinitely many circles through A, B which lie entirely inside C.

Solution

Let O be the center of the circle C. The perpendicular bisector of AB must meet the side OA or the side OB. wlog let it meet the side OB. If P is any point on the segment OB, then the circle center P radius PB certainly lies entirely inside C. So this is true in particular if P lies on the perpendicular bisector of AB. But in this case B also lies on the small circle. So we have found one circle.

But if we now take any point on the perpendicular bisector sufficiently close to P then the circle center P through A (and hence also B) will also lie entirely inside C.