9.5. Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the sum of the numbers written next to its row and its column (a "sum table"). What is the maximum number of sums in this table that could be rational numbers?

(O. Podlipsky)
Answer: 1250 sums.

Solution. First, we show that there are at least 1250 irrational numbers in the table. Let there be $x$ irrational and $50-x$ rational numbers along the left side of the table. Then, along the top side, there are $50-x$ irrational and $x$ rational numbers. Since the sum of a rational and an irrational number is always irrational, there are at least $x^2 + (50-x)^2$ irrational numbers in the table. Furthermore, $x^2 + (50-x)^2 = 2x^2 - 100x + 50^2 = 2(x-25)^2 + 2 \cdot 25^2 \geq 2 \cdot 25^2 = 1250$, which is what we needed to show. Therefore, there are no more than $2500 - 1250 = 1250$ rational numbers in the table.

Exactly 1250 rational numbers in the table can occur, for example, in the following case. Along the left side, the numbers are $1, 2, \ldots, 24, 25, 1+\sqrt{2}, 2+\sqrt{2}, \ldots, 25+\sqrt{2}$, and along the top side, the numbers are $26, 27, \ldots, 49, 50, 26-\sqrt{2}, 27-\sqrt{2}, \ldots, 50-\sqrt{2}$. Then, the only irrational sums will be $2 \cdot 25^2 = 1250$ sums of a rational and an irrational number.

Comment. Only the correct answer - 0 points.

Proof that there are no more than 1250 rational numbers in the table - 4 points.

If this proof is only conducted in a special case where there are an equal number of rational and irrational numbers along the left (or top) side - 1 point instead of four.
The inequality $x^2 + (50-x)^2 \geq 1250$ (or an equivalent one) is used without proof - 1 point is deducted.

A correct example showing that there could be exactly 1250 rational numbers - 3 points.

The statement that the sum of a rational and an irrational number is irrational can be used without proof. 9.6. In an acute triangle $ABC$, the median $AM$ and the altitude $BH$ are drawn. The perpendicular from point $M$ to line $AM$ intersects ray $HB$ at point $K$. Prove that if $\angle MAC = 30^\circ$, then $AK = BC$.

(B. Obukhov)

First Solution. Since $\angle AHK = \angle AMK = 90^\circ$, points $A, H, M$, and $K$ lie on a circle $\omega$ with diameter $AK$ (see Fig. 1). By the problem's condition, the chord $HM$ of this circle subtends an angle $\angle MAH = 30^\circ$, so $HM = 2R \sin 30^\circ = AK / 2$. On the other hand, $HM$ is the median of the right triangle $BHC$, so $BC = 2HM = AK$, which is what we needed to prove.

![](https://cdn.mathpix.com/cropped/2024_05_06_4f9f6358f594c2adc161g-1.jpg?height=358&width=421&top_left_y=988&top_left_x=1691)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_4f9f6358f594c2adc161g-1.jpg?height=485&width=604&top_left_y=928&top_left_x=2140)

Fig. 2
Second Solution. Extend triangle $ABC$ to parallelograms $ABPC$ and $AQB C$ (see Fig. 2); then $BP = AC = BQ$ and $KH \perp PQ$. Moreover, point $M$ is the midpoint of $AP$ (as the intersection of the diagonals of parallelogram $ABPC$). Therefore, lines $MK$ and $HK$ are the perpendicular bisectors of segments $AP$ and $PQ$. Consequently, point $K$ is the center of the circle $\Omega$ circumscribed around triangle $APQ$. Further, $\angle APQ = \angle PAC = 30^\circ$, so the chord $AQ$ of circle $\Omega$ equals the radius $AK$. Finally, from parallelogram $AQB C$, we get $BC = AQ = AK$.

Comment. Noted that points $A, H, M$, and $K$ lie on the same circle - 2 points.