Problem 1325

I. Solution: Connect the points where the angle sides intersect the circle. Draw a diameter parallel to this chord and draw lines parallel to the angle sides through the endpoints of this diameter.

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These parallel lines must intersect each other on the circle, because the geometric locus of points from which a given distance is seen at a right angle is the circle drawn over that distance as a diameter. The resulting triangles are similar, and the larger one has a hypotenuse that is a diameter of the circle, so the sum of the legs is also larger in this case. For every right angle in the circle, we can find a corresponding inscribed angle whose sides are segments of the circle, and the sum of these segments is larger. Therefore, it is sufficient to look among the right triangles drawn over the diameter to find the one with the largest sum of legs. We will show that this follows for the isosceles right triangle. Let $ABC$ be the isosceles triangle and $ABC_1$ be any right triangle over the diameter $AB$.

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Reflect the side $BC_1$ over the external angle bisector through $C_1$. Let the reflection be $C_1B'$. After reflecting over the external angle bisector, $A$, $C_1$, and $B'$ lie on the same line. The external angle bisector forms a $45^\circ$ angle with the legs. Furthermore, $\angle AC_1C = \angle ABC = 45^\circ$ because they are inscribed angles subtending the same arc. Thus, the external angle bisector also passes through $C$, so the reflection of $BC$ is $B'C$. From $\triangle AB'C$, we have $AC + CB = AC + CB' > AB' = AC_1 + C_1B' = AC_1 + C_1B$.

This proves that the sum of the segments of the sides of the right angle is the largest when the sides pass through the endpoints of a diameter and the vertex is the center of the semicircle above this diameter.

II. Solution: Let the lengths of the segments of the angle sides that lie on the circle be $a$ and $b$. For any two positive numbers $a$ and $b$, the inequality

$$\frac{a+b}{2} \leq \sqrt{\frac{a^2 + b^2}{2}} = \frac{\sqrt{a^2 + b^2}}{\sqrt{2}}$$

holds, and equality holds if and only if $a = b$.

Before proving this, let's see how it helps solve the problem. If $a$ and $b$ are the segments in question, $a + b$ and thus $\frac{a + b}{2}$ are maximized when the right-hand side is maximized and the left-hand side is equal to it, assuming this case can occur. The geometric meaning of the numerator on the right-hand side is the length of the chord connecting the points where the angle sides intersect the circle. This is maximized when this chord is a diameter. In this case, the left-hand side is equal to the right-hand side if and only if $a = b$, i.e., when the right triangle is isosceles with the hypotenuse as the diameter.

Our solution will be complete if we prove the inequality (1). It suffices to prove the corresponding inequality between the squares of the two sides, since both sides are positive and the larger of two positive numbers is the one whose square is larger. Therefore, we will prove that

$$\left(\frac{a+b}{2}\right)^2 \leq \frac{a^2 + b^2}{2}$$

We show that the difference between the two sides cannot be negative:

$$\frac{a^2 + b^2}{2} - \left(\frac{a+b}{2}\right)^2 = \frac{2a^2 + 2b^2 - a^2 - 2ab - b^2}{4} = \frac{a^2 - 2ab + b^2}{4} = \left(\frac{a-b}{2}\right)^2 \geq 0$$

and equality holds if and only if $a = b$. This implies that the above equality, and thus the inequality (1), holds, and equality holds in these cases if and only if $a = b$.

III. Solution: Draw any broken line whose segments form the same angle with a given line.

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It is clear that the total length of this broken line is the longest when its projection on the line is the longest. (Of course, if the projections of multiple segments of the broken line fall on the same segment of the line, we count this segment multiple times in the projection.)

Draw the external angle bisector of the given right angle. This bisector forms the same $45^\circ$ angle with both sides of the angle. Project the sides of the right angle onto this bisector. This projection is the same as the projection of the chord that the sides of the angle cut out from the circle. The projection cannot be longer than the diameter of the circle. It will be equal to the diameter if the right angle lies on the diameter and this diameter is parallel to the drawn external angle bisector. This follows if the diameter and the sides of the angle form an isosceles right triangle.

Remark: This also provides a simple geometric proof of the inequality used in the previous solution.

[^0]: ${ }^{1} \mathrm{The}$ inequality follows from the inequality proved in problem $236^{*}$ by substituting $c=d=1$.