Problem 972

I. Solution: a) The center $K$ of the circumscribed circle can be constructed using an angle. If one of the angles of the triangle $ABC$, for example, $\angle ACB = \gamma$ is a right angle, then $K$ coincides with the midpoint $C_0$ of the opposite side $AB$, and can be constructed by drawing a parallelogram with $AB$ as the diagonal and intersecting $AB$ with the other diagonal. If $\gamma$ is an acute angle, then the angle subtended by $AB$ from $K$ is $2\gamma$, and $K$ is on the same side of $AB$ as $C$. The triangle $ABK$ is isosceles, so $\angle KAB = \angle KBA = (180^\circ - 2\gamma) / 2 = 90^\circ - \gamma$; based on this, $K$ can be constructed. If $\gamma$ is an obtuse angle, then the angle subtended by $AB$ from $K$ is $360^\circ - 2\gamma$, so $\angle KAB = \angle KBA = \gamma - 90^\circ$ and $K$ must be constructed on the side of $AB$ opposite to $C$.

b) The center $O$ of the inscribed circle $k$, and the center $O_c$ of the excircle $k_c$ opposite to side $AB = c$ can be constructed using the sides as follows: $k$ touches the sides $CA$ and $CB$ at a distance of $s - c = (a + b - c) / 2$ from $C$ towards $A$ and $B$, respectively, and $k_c$ touches the extensions of these sides at a distance of $s = (b + c + a) / 2$ from $C$ beyond $A$ and $B$. The intersection of the perpendiculars to the corresponding sides at the two points of tangency gives $O$ and $O_c$. Similarly, $O_a$ and $O_b$ can be constructed.

c) The centroid $S$ can be obtained, for example, by constructing the median $s_c$ to side $AB$ in any triangle, as described for the case of a right triangle in part a). Similarly constructing $s_b$, it intersects $s_c$ at $S$.

Julianna Egyed (Gyöngyös, Vak Bottyán g. I. o. t.)

Remarks. 1. The construction in part a) can be stated simultaneously for acute and obtuse angles: perpendicular rays $a^*$ and $b^*$ are drawn from $A$ and $B$ in the half-plane containing $C$, and $\gamma$ is measured from the side on which segment $AB$ lies.

2. It may be noted that we have used a line that is also the bisector of a side or angle. If this is not allowed, even in a different construction, we must examine whether we have implicitly used a side or angle bisector. In part a), the lines $AK$ and $BK$ become both a side and an angle bisector if the sides from $A$ and $B$ are equal. Therefore, the given procedure is not allowed in an equilateral triangle and can only be used in an isosceles triangle when $AB \neq BC = CA$. - In part b), the perpendicular at the point of tangency of $k$ cannot be used if the point of tangency bisects the side in question. The point $O$ (and similarly $S$) can only be constructed with the given method when $AB \neq BC = CA$.
3. The point $K \equiv O \equiv S$ of an equilateral triangle can be obtained, for example, by taking points $C_1$, $A_1$, and $B_1$ on the extensions of sides $AB$, $BC$, and $CA$ beyond $B$, $C$, and $A$ such that $C_1B = A_1C = B_1A = AB$, and constructing the center of the equilateral triangle $A_1B_1C_1$.

II. Solution: a) From the two known formulas for the area of the triangle, $t = cm_c / 2 = abc / 4r$, and from this $KA = r = ab / 2m_c$, which can be constructed as a fourth proportional, and the common point of the circles of radius $r$ drawn from the vertices around $r$ is $K$ (all three circles are necessary).

b) Similarly, from $2t = cm_c = 2\rho s = 2\rho_c(s - c)$, we get $\rho = cm_c / 2s$, $\rho_c = cm_c / 2(s - c)$, so the intersection of the parallels at a distance of $\rho$ and $\rho_c$ from the lines $CA$ and $CB$ on the side containing $B$ and $A$, respectively, is $O$ and $O_c$.

c) Similarly, the intersection of the parallels through the points that divide the corresponding heights into thirds gives $S$.

Gábor Fábián (Győr, Bencés g. II. o. t.)

Remarks. 1. If two sides are equal, the height from their common vertex cannot be used for the above purpose, and if all three sides are equal, this solution is also unusable.

2. If the sides are different, the construction of $K$ can use the fact that the reflections of the orthocenter $M$ over the sides lie on the circumcircle. In this case, all three heights and all three reflections can be constructed, and the center of the circle passing through the latter is $K$.

Gábor Popper (Budapest, Bolyai J. g. I. o. t.)

3. The perpendiculars from the vertices to the sides of the pedal triangle cut out a diameter from the circumcircle; $K$ can also be constructed based on this.

Béla Bollobás (Budapest, Apáczai Csere J. gyak. g. II. o. t.)