In the right-angled triangle $ABC$, at some point $D$ on the leg $BC$, the perpendicular to $AD$ intersects the hypotenuse $AB$ at $E$, and the perpendicular from $E$ to $BC$ meets $BC$ at $F$. For which choice of point $D$ will $F$ be closest to $C$?
I. solution. Let the orthogonal projection of $E$ onto $AC$ be $E_{1}$, then $E_{1}E = CF$ and $\triangle A E_{1} E \sim \triangle A C B$, so $CF$ is the smallest when $AE$ is the smallest.

![](https://cdn.mathpix.com/cropped/2024_05_02_ef1fb532cd417649deceg-1.jpg?height=597&width=637&top_left_y=268&top_left_x=728)

Figure 1

The semicircle drawn over $AE$ as a diameter passes through point $D$ on side $CB$ and vice versa, if a semicircle with diameter $AE$ on $AB$ passes through a point $D$ on $BC$, then $DE \perp AD$, and thus $E$ and the projection of $F$ on $CB$ are the points formed as described in the problem.

Thus, we reduce the problem to the following: find the semicircle with the smallest diameter, which has $AE$ as its diameter on line $AB$ and has a common point with $BC$. We will show that this is the semicircle that is tangent to $BC$. Indeed, draw the tangent to $BC$ parallel to a semicircle with the described properties. This tangent, together with lines $AB$ and $AC$, forms a triangle similar to and containing $\triangle ABC$; thus, by scaling down, we get $\triangle ABC$ and the semicircle tangent to $BC$. The diameter of this semicircle is therefore the smallest among the semicircles we are examining, as we claimed. Let the center of this semicircle be $K'$, the point of tangency $D'$, then $BC \perp K'D' \parallel AC$, and $AD'K'$ is an isosceles triangle, so

$$
\angle K'AD' = \angle K'D'A = \angle D'AC,
$$

thus the point $D$ that leads to the shortest $CF$ and the shortest $AE$ is the intersection of the angle bisector of $\angle CAB$ with the leg $CB$.

II. solution. Due to the construction, $\triangle DEF$ and $\triangle ADC$ are similar,

$$
DF = AC \cdot \frac{DE}{AD}
$$

so the length of $DF$ can also be obtained by scaling down the right triangle $\triangle ADE$, taking the leg starting from $A$ to be of length $AC$, in which case the desired length is the other leg. This can be simply obtained by measuring $AC$ on line $AB$ and intersecting the perpendicular at the endpoint $C_1$ with line $AD$ at $D_1$; thus $C_1D_1 = DF$, and $CF = CD + C_1D_1$. Then the line $C_1D_1$ is the reflection of $CB$ over the angle bisector of $\angle CAB$, so the intersection point $G$ of the lines $CB$ and $C_1D_1$ lies on the angle bisector.

![](https://cdn.mathpix.com/cropped/2024_05_02_ef1fb532cd417649deceg-1.jpg?height=597&width=520&top_left_y=1981&top_left_x=781)

Figure 2

If $D$ is taken exactly at $G$, then $D_1$ also appears at $G$, and $CF = CD + DF = CG + C_1D_1 = CG + C_1G = 2 \cdot CG$. We will show that in any other position of $D$, $CF > 2CG$, i.e., the desired position of $D$ is exactly the point $G$. The angle $C D G = 90^\circ - \angle C_1 A D_1 = \angle A D_1 C_1 = \angle D D_1 G$,

so $D_1G > DG$, hence $CF = CD + C_1G + GD_1 > CD + C_1G + DG = CG + C_1G = 2CG$, which we wanted to prove.

Csóka Géza (Ajka, Bródy I. g. III. o. t.) Bottyán István (Hatvan, Bajza J. g. III. o. t.)

III. solution. $\triangle DEF$ and $\triangle ADC$, as well as $\triangle EFB$ and $\triangle ACB$ are similar right triangles (Figure 1), so with $AC = b$, $CB = a$, $CD = x$, $CF = y$, and $AB = c$, we have on the one hand

$$
\begin{aligned}
& DF = EF \cdot \frac{b}{x}, \quad y = x + \frac{b}{x} \cdot EF, \text{ and on the other hand } \\
& \frac{EF}{b} = \frac{FB}{a} = \frac{a - y}{a}, \quad EF = b - \frac{b}{a} \cdot y, \quad \text{ so } \quad y = x + \frac{b^2}{x} - \frac{b^2}{ax} \cdot y \\
& y = a \cdot \frac{x^2 + b^2}{ax + b^2}
\end{aligned}
$$

Removing the fraction, reducing the equation to 0, and completing the square on the left side:

$$
\begin{gathered}
0 = ax^2 - axy + b^2(a - y) = a\left(x - \frac{y}{2}\right)^2 + b^2(a - y) - a \frac{y^2}{4} \\
a\left(x - \frac{y}{2}\right)^2 = \frac{a}{4} \cdot y^2 + b^2 y - ab^2 = a\left[\left(\frac{y}{2} + \frac{b^2}{a}\right)^2 - b^2 \cdot \frac{a^2 + b^2}{a^2}\right] = \\
= a\left[\left(\frac{y}{2} + \frac{b^2}{a}\right)^2 - \frac{b^2 c^2}{a^2}\right]
\end{gathered}
$$

Here, the letters represent positive distances, so the left side is non-negative, and thus the right side cannot be negative. The value of $y$ is the smallest when the first term in the brackets is, i.e., when

$$
\frac{y}{2} + \frac{b^2}{a} = \frac{bc}{a}, \quad y = \frac{2b(c - b)}{a}
$$

Then

$$
x = \frac{y}{2} = \frac{b(c - b)}{a} = a \cdot \frac{b(c - b)}{a^2} = a \cdot \frac{b(c - b)}{c^2 - b^2} = a \cdot \frac{b}{c + b}
$$

which we recognize as the segment cut by the angle bisector.

Külvári István (Budapest, Széchenyi I. g. III. o. t.)

Remark. The above can also be obtained through trigonometric means. Let $AC = 1$, $\angle CAD = \varphi$, $\angle CAB = \alpha$, so $\angle DAB = \alpha - \varphi$, and $CD = \tan \varphi$, $AD = 1 / \cos \varphi$, $DE = AD \tan(\alpha - \varphi)$, $DF = DE \cos \varphi = \tan(\alpha - \varphi)$, and using the addition theorem, converting the product in the denominator to a sum, $\cos u \cos v = [\cos(u + v) + \cos(u - v)] / 2$, and finally increasing the denominator,

$$
\begin{gathered}
CF = \frac{\sin \varphi}{\cos \varphi} + \frac{\sin(\alpha - \varphi)}{\cos(\alpha - \varphi)} = \frac{\sin(\varphi + \alpha - \varphi)}{\cos \varphi \cos(\alpha - \varphi)} = \frac{2 \sin \alpha}{\cos(\varphi + \alpha - \varphi) + \cos(2\varphi - \alpha)} \geq \\
\geq \frac{2 \sin \alpha}{\cos \alpha + 1} = 2 \sqrt{\frac{1 - \cos^2 \alpha}{(1 + \cos \alpha)^2}} = 2 \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} = 2 \tan \frac{\alpha}{2}
\end{gathered}
$$

and equality holds only when $\cos(2\varphi - \alpha) = 1$, $2\varphi - \alpha = 0$, $\varphi = \alpha / 2$.

Similarly, somewhat more simply,

$$
AE = \frac{AD}{\cos(\alpha - \varphi)} = \frac{1}{\cos \varphi \cos(\alpha - \varphi)} \geq \frac{2}{\cos \alpha + 1} = \frac{1}{\cos^2 \alpha / 2}
$$

Karsai István (Szeged, Radnóti M. g. III. o. t.)