Given the sides of triangle $A B C$ are $a, b, c$. The triangle, where $b>c$, is divided into two equal areas by the line $H K$ perpendicular to the base $B C$. What is the distance from the foot of this perpendicular $H K$ to $C$?
According to the problem:

$$
a b = 2 C K \cdot C H
$$

but

$$
C H = \frac{C K}{\cos \gamma}
$$

thus

$$
a b = \frac{2 \cdot C K^2}{\cos \gamma}
$$

from which

$$
C K = \sqrt{\frac{a b}{2} \cos \gamma}
$$

By the cosine rule:

$$
c^2 = a^2 + b^2 - 2 a b \cos \gamma
$$

from which

$$
\frac{a b \cos \gamma}{2} = \frac{a^2 + b^2 - c^2}{4}
$$

Substituting (2) into (1):

$$
C K = \frac{1}{2} \sqrt{a^2 + b^2 - c^2}
$$

Construction. (1) can also be written as:

$$
C K = \sqrt{\frac{a}{2} \cdot C D}
$$

Thus, \( C K \) is the geometric mean between \( \frac{a}{2} \) and \( C D \). Therefore, a perpendicular is drawn at the midpoint \( E \) of side \( B C \); a semicircle is drawn over \( D C \), which intersects \( E F \) at point \( G \). \( C G = C K \) will be the desired distance.

(Thiringer Aurél, secondary school VII. grade, Sopron.)

The 273rd problem was solved by: Bálint Béla, Dénes Aladár, Eislitzer Gyula, Frankl Ignácz, Freund Antal, Friedmann Bernát, Geist Emil, Goldstein Zsigmond, Goldzieher Károly, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Kornis Ödön, Kreizler Győző, Piovarcsy Jenő, Preisz Károly, Riesz Frigyes, Roth Miksa, Schiffer Hugó, Schneider Béla, Spitzer Ödön, Szabó István, Szabó Károly, Szigeth Gábor, Weisz Ármin, Weisz Lipót.

The 274th problem was solved by: Freund Antal, Friedmann Bernát, Geist Emil, Goldstein Zsigmond, Goldzieher Károly, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Kornis Ödön, Riesz Frigyes, Roth Miksa, Spitzer Ödön, Szabó István, Szabó Károly, Szigeth Gábor, Weisz Ármin, Weisz Lipót.