In a certain triangle $ABC$, from the points on side $AB = c$, we draw perpendiculars to the other two sides. Determine the point $P$ on $c$ such that the sum of the squares of the perpendiculars drawn from it is minimized.
I. solution: The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_a862fc9dd0b2d6b1f137g-1.jpg?height=448&width=526&top_left_y=226&top_left_x=775)

Figure 1

If $AP = x$, then $BP = c - x$, and the two perpendiculars are $u = x \sin \alpha, v = (c - x) \sin \beta$.

Thus,

$$
u^2 + v^2 = x^2 \sin^2 \alpha + (c - x)^2 \sin^2 \beta = x^2 (\sin^2 \alpha + \sin^2 \beta) - 2cx \sin^2 \beta + c^2 \sin^2 \beta
$$

is a quadratic function, and we are looking for its minimum.

It is known that the function $y = ax^2 + bx + c$ takes its extreme value at $x = -\frac{b}{a}$, and this extreme value is a minimum if $a > 0$.

In this case, since $\sin^2 \alpha + \sin^2 \beta > 0$, the minimum occurs at

$$
x = \frac{c \sin^2 \beta}{\sin^2 \alpha + \sin^2 \beta}
$$

and thus

$$
c - x = \frac{c \sin^2 \alpha}{\sin^2 \alpha + \sin^2 \beta}
$$

Using the sine rule,

$$
AP : PB = x : (c - x) = \sin^2 \beta : \sin^2 \alpha = b^2 : a^2
$$

Deseő Zoltán (Bp. X., I. László g. IV. o. t.)

Remark: The calculated ratio allows the point $P$ to be easily constructed. For example,

$$
b^2 : a^2 = b : \frac{a^2}{b} = b : z, \quad \text{where} \quad b : a = a : z
$$

The segment $z$ can thus be obtained as the fourth proportional. The construction can be read from Figure 1.

II. solution: The point $P$ can also be constructed purely geometrically without calculations.

Take any point $P_x$ on the side $AB$. Let the feet of the perpendiculars from $P_x$ be $Q_x$ and $R_x$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_a862fc9dd0b2d6b1f137g-1.jpg?height=406&width=482&top_left_y=2071&top_left_x=824)

Figure 2

Rotate the segment $P_x Q_x = u$ around $P_x$ to the position $P_x Q_x'$, where $P_x Q_x' \perp P_x R_x$.

As $P_x$ runs along the side $AB$, the isosceles triangles $P_x Q_x Q_x'$ are all similar and perspective from the center $A$. Therefore, the locus of the points $Q_x'$ is the line $AQ_x'$, which intersects the side $BC$ at the point $C'$, where $BC'$ is clearly the altitude $m_b$ from $B$ rotated around $P \equiv B$. By the Pythagorean theorem, $u^2 + v^2 = Q_x' R_x^2$. The segment $Q_x' R_x$ is the diagonal of the rectangle $P_x Q_x' S_x R_x$, which is inscribed in the triangle $ABC'$ such that the side $R_x S_x$ lies on the side $BC'$ of the triangle.

Our problem can now be stated as: Among the rectangles inscribed in the triangle $ABC'$ (one side of which lies on the segment of the side $BC'$), which one has the smallest diagonal?

To determine this, translate the vertex $A$ parallel to the side $BC'$ to an arbitrary point $A_1$ (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_a862fc9dd0b2d6b1f137g-2.jpg?height=442&width=569&top_left_y=587&top_left_x=737)

The rectangle $DEFG$ inscribed in the triangle $ABC'$ is then translated to the rectangle $D_1E_1F_1G_1$ inscribed in the triangle $A_1CB$, which is congruent to the original rectangle because $G_1F_1 : BC' = A_1G_1 : A_1B = AG : AB = GF : BC'$, and thus

$$
G_1F_1 = GF.
$$

If the point $A$ is translated to the point $A_2$, where $A_2C' \perp C'B$, then our problem can be stated as: Among the rectangles inscribed in the right triangle $BC'A_2$ (one vertex of which is the point $C'$), which one has the smallest diagonal? The distance from the point $P_2$ on the hypotenuse $A_2B$ to the point $C'$ is the diagonal of the rectangle, which is clearly minimized when $C'P_2 \perp A_2B$.

The construction process is as follows: Construct the point $C'$ on the side $BC$ such that $BC' = m_b$ (Figure 2). Draw a perpendicular from the point $C'$ to the side $BC'$ and measure the distance $C'A_2 = m_a$ (Figure 3). Draw a perpendicular from $C'$ to the hypotenuse $A_2B$, and let its foot be $P_2$. Draw a line through $P_2$ parallel to $BC'$, and let it intersect the side $AB$ at the desired point $P$.

Szendrei István (Kunszentmiklós, Damjanich g. III. o. t.)

Remark: It is easy to verify that the point $P$ constructed in this way corresponds to the point $P$ obtained in the first solution. Indeed, in the right triangle $A_2C'B$ (Figure 3), by a known theorem,

$$
\begin{aligned}
& A_2C'^2 = m_a^2 = A_2B \cdot A_2P_2 \\
& BC'^2 = m_b^2 = A_2B \cdot P_2B
\end{aligned}
$$

Thus,

$$
\frac{m_a^2}{m_b^2} = \frac{A_2P_2}{P_2B} = \frac{AP}{PB}
$$

By the area formula, $\frac{m_a}{m_b} = \frac{b}{a}$, and thus $\frac{AP}{PB} = \frac{b^2}{a^2}$, which agrees with the result obtained in the first solution.