Kvant Math Problem 276

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m33s
Source on kvant.digital

Problem

Given a square $ABCD$. Points $P$ and $Q$ lie respectively on sides $AB$ and $BC$, and $BP = BQ$. Let $H$ be the foot of the perpendicular dropped from point $B$ to segment $PC$. Prove that the angle $DHQ$ is a right angle.

N. B. Vasiliev

All-Union Mathematical Olympiad for School Students (1974, 10th grade)

Exploration

A direct synthetic approach would require tracking the foot of a perpendicular from $B$ to the line $PC$, which suggests that a coordinate representation or vector projection will likely reduce the configuration to an algebraic identity. The condition $BP = BQ$ indicates a symmetry between the two legs $AB$ and $BC$, suggesting that $P$ and $Q$ can be parameterized by a single variable along perpendicular axes.

Choosing a Cartesian model with the square aligned to coordinate axes converts the geometric condition into a single parameter $t$ describing both $P$ and $Q$. The foot $H$ is then obtained as an orthogonal projection of the origin onto a line through two explicitly known points, which is stable under computation. The final condition $\angle DHQ = 90^\circ$ translates into a dot product between vectors $HD$ and $HQ$ equal to zero.

The main risk lies in algebraic simplification of the projection point and in ensuring that no computational cancellation errors hide a nontrivial geometric dependency. Once the projection is correctly expressed, the orthogonality condition reduces to a polynomial identity in one variable.

Problem Understanding

The figure is a square $ABCD$ with $P$ on $AB$ and $Q$ on $BC$, chosen so that $BP = BQ$. From $B$, a perpendicular is dropped to the segment $PC$, meeting it at $H$. The task is to prove that $\angle DHQ$ is a right angle.

This is a Type B problem, requiring proof of a geometric orthogonality statement.

The constraint $BP = BQ$ enforces a balanced placement of $P$ and $Q$ on two perpendicular sides, suggesting hidden rotational or reflective symmetry centered at $B$. The conclusion claims that the segment $HD$ is perpendicular to $HQ$, even though $D$ lies diagonally opposite $B$ and $Q$ lies on an adjacent side, so the structure is not immediately symmetric in the obvious axes.

Proof Architecture

Introduce a coordinate system placing the square in the plane so that $B$ is at the origin, $A$ and $C$ lie on coordinate axes, and $D$ is determined accordingly. Parameterize $P$ and $Q$ using a single parameter $t$ imposed by the condition $BP = BQ$.

Compute the equation of the line $PC$ and determine $H$ as the orthogonal projection of the origin onto this line by minimizing the squared distance to a general point on $PC$. This yields an explicit formula for $H$.

Express vectors $HD$ and $HQ$ in coordinates and compute their scalar product. Show that this scalar product vanishes identically in $t$, establishing orthogonality.

The most delicate point is the correct computation of the projection parameter and the simplification of the resulting coordinates of $H$, since any algebraic error would destroy the cancellation leading to the final identity.

Solution

Place the square so that $B = (0,0)$, $A = (0,1)$, $C = (1,0)$, and $D = (1,1)$. Let $P$ lie on $AB$, so $P = (0,t)$ for some $t \in [0,1]$. Let $Q$ lie on $BC$, so $Q = (t,0)$ since the condition $BP = BQ$ implies that both distances from $B$ are equal to $t$.

The line $PC$ passes through $P = (0,t)$ and $C = (1,0)$, hence it can be parameterized as

$$X(s) = (s,, t - st), \quad s \in \mathbb{R}.$$

The point $H$ is the foot of the perpendicular from the origin to this line, so it minimizes

$$|X(s)|^2 = s^2 + (t - st)^2.$$

Expanding,

$$|X(s)|^2 = s^2 + t^2 - 2st^2 + s^2 t^2 = s^2(1+t^2) - 2st^2 + t^2.$$

Differentiating with respect to $s$ and setting the derivative equal to zero yields

$$2s(1+t^2) - 2t^2 = 0,$$

so

$$s = \frac{t^2}{1+t^2}.$$

Substituting this value gives the coordinates of $H$:

$$H = \left(\frac{t^2}{1+t^2},, t - \frac{t^3}{1+t^2}\right) = \left(\frac{t^2}{1+t^2},, \frac{t}{1+t^2}\right).$$

The vectors $HD$ and $HQ$ are computed as follows. For $D = (1,1)$,

$$HD = \left(1 - \frac{t^2}{1+t^2},, 1 - \frac{t}{1+t^2}\right) = \left(\frac{1}{1+t^2},, \frac{1+t^2 - t}{1+t^2}\right).$$

For $Q = (t,0)$,

$$HQ = \left(t - \frac{t^2}{1+t^2},, -\frac{t}{1+t^2}\right) = \left(\frac{t + t^3 - t^2}{1+t^2},, -\frac{t}{1+t^2}\right).$$

The scalar product satisfies

$$HD \cdot HQ = \frac{1}{(1+t^2)^2}\Bigl[(t + t^3 - t^2) - t(1+t^2 - t)\Bigr].$$

Expanding the expression in brackets,

$$t + t^3 - t^2 - t - t^3 + t^2 = 0.$$

Hence $HD \cdot HQ = 0$, which implies $HD \perp HQ$, so $\angle DHQ = 90^\circ$.

This completes the proof. ∎

Verification of Key Steps

The computation of $H$ relies on minimizing the squared distance from the origin to a parametrized point on $PC$. The resulting linear equation in $s$ has a unique solution because the coefficient $1+t^2$ is strictly positive, ensuring that the projection is well-defined and unique.

The simplification of the second coordinate of $H$ uses the identity

$$t - \frac{t^3}{1+t^2} = \frac{t(1+t^2) - t^3}{1+t^2} = \frac{t}{1+t^2},$$

which confirms that no hidden dependence on higher powers of $t$ remains.

In the scalar product computation, every term in $t$, $t^2$, and $t^3$ cancels exactly between the contributions from $HD$ and $HQ$, and the cancellation does not rely on any inequality or approximation, only algebraic rearrangement.

Alternative Approaches

A synthetic approach can be built using a rotation of $90^\circ$ about $B$, sending $A$ to $C$ and $P$ to $Q$. Under this transformation, the line $PC$ maps to itself, and the foot of the perpendicular from $B$ to $PC$ is preserved. The configuration then exhibits a symmetry exchanging $D$ with a point whose alignment with $H$ and $Q$ encodes the right angle condition.

The coordinate method is more direct because it converts the geometric projection into a single explicit minimization problem and reduces the final statement to an algebraic identity, avoiding the need to manage rotational invariants geometrically.