Kvant Math Problem 802

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Verdicts: SKIP + SKIP
Solve time: 2m14s
Source on kvant.digital

Problem

On the sides $AB$ and $BC$ of triangle $ABC$, right triangles $APB$ and $BQC$ are constructed externally using these sides as hypotenuses, with identical angles of measure $\beta$ at their common vertex $B$ (see Fig. 1). Find the angles of triangle $PQK$, where $K$ is the midpoint of side $AC$.

Fig. 1

L. P. Kuptsov

Exploration

The data involve two right triangles erected externally on sides $AB$ and $BC$. The angle at the common vertex $B$ in each right triangle equals $\beta$. Since $AB$ and $BC$ are the hypotenuses, the right angles are at $P$ and $Q$ respectively.

A natural coordinate model is to place

$$B=(0,0),\qquad A=(a,0),\qquad C=(c_x,c_y).$$

Because $\angle ABP=\beta$ and $\angle BQC=\beta$, the points $P$ and $Q$ are obtained from $A$ and $C$ by multiplying the vectors $BA$ and $BC$ by $\cos\beta$ and rotating them by $\pm\beta$.

For the first triangle,

$$P=(a\cos^2\beta,,-a\sin\beta\cos\beta).$$

For the second,

$$Q=(c_x\cos^2\beta-c_y\sin\beta\cos\beta,, c_x\sin\beta\cos\beta+c_y\cos^2\beta).$$

The midpoint of $AC$ is

$$K=\frac{A+C}{2}.$$

The expressions for $P,Q,K$ suggest computing

$$\overrightarrow{KP},\qquad \overrightarrow{KQ}.$$

After simplification,

$$\overrightarrow{KP} = -\frac12 \begin{pmatrix} \cos 2\beta & \sin 2\beta\ -\sin 2\beta & \cos 2\beta \end{pmatrix} (C-A),$$

while

$$\overrightarrow{KQ} = \frac12 \begin{pmatrix} \cos 2\beta & -\sin 2\beta\ \sin 2\beta & \cos 2\beta \end{pmatrix} (C-A).$$

These matrices are rotations by $-2\beta$ and $2\beta$. Hence

$$|KP|=|KQ|=\frac12,|AC|.$$

Thus $K$ is on the perpendicular bisector of $PQ$, and triangle $PQK$ is isosceles.

The angle between $\overrightarrow{KP}$ and $\overrightarrow{KQ}$ equals the angle between

$$-R_{-2\beta}(C-A) \quad\text{and}\quad R_{2\beta}(C-A).$$

The minus sign contributes a rotation by $\pi$, so the angle is

$$\pi-4\beta.$$

Therefore

$$\angle PKQ=180^\circ-4\beta,$$

and the equal base angles are

$$\frac{180^\circ-(180^\circ-4\beta)}2=2\beta.$$

The step most likely to conceal an error is the derivation of the vector formulas for $\overrightarrow{KP}$ and $\overrightarrow{KQ}$, because a sign mistake in the rotations would change the final angle.

Problem Understanding

We are given a triangle $ABC$. On the sides $AB$ and $BC$, external right triangles $APB$ and $BQC$ are constructed. Their hypotenuses are $AB$ and $BC$ respectively, and the angles at the common vertex $B$ are both equal to $\beta$.

Let $K$ be the midpoint of $AC$. We must determine the three angles of triangle $PQK$.

This is a Type C problem, since the required quantities are specific values to be determined.

The core difficulty is to relate the geometry of the auxiliary right triangles to the midpoint of $AC$. The computation in the exploration suggests that $K$ is equidistant from $P$ and $Q$, and that the angle at $K$ depends only on $\beta$, not on the shape of triangle $ABC$.

The expected answer is

$$\angle KPQ=\angle KQP=2\beta,\qquad \angle PKQ=180^\circ-4\beta.$$

The reason is that the vectors from $K$ to $P$ and $Q$ turn out to be rotated copies of the same half-diagonal vector.

Proof Architecture

Lemma 1. If $v=\overrightarrow{BA}$, then $\overrightarrow{BP}$ is obtained from $v$ by multiplication by $\cos\beta$ and rotation through $-\beta$; this follows from the right triangle $APB$ with hypotenuse $AB$.

Lemma 2. If $w=\overrightarrow{BC}$, then $\overrightarrow{BQ}$ is obtained from $w$ by multiplication by $\cos\beta$ and rotation through $+\beta$; this follows analogously from triangle $BQC$.

Lemma 3. Writing $R_\theta$ for rotation through angle $\theta$,

$$\overrightarrow{KP} = -\frac12R_{-2\beta}(C-A), \qquad \overrightarrow{KQ} = \frac12R_{2\beta}(C-A).$$

This is obtained by substituting the formulas from Lemmas 1 and 2 and simplifying.

Lemma 4. The lengths $KP$ and $KQ$ are equal and

$$\angle PKQ=180^\circ-4\beta.$$

This follows because rotations preserve length and because the angle between the vectors in Lemma 3 is $\pi-4\beta$.

The hardest point is Lemma 3. Any sign error in the rotation formulas changes the final result.

Solution

Let $R_\theta$ denote rotation in the plane through angle $\theta$.

Put

$$v=\overrightarrow{BA},\qquad w=\overrightarrow{BC}.$$

Since triangle $APB$ is right-angled at $P$, has hypotenuse $AB$, and satisfies

$$\angle ABP=\beta,$$

the vector $\overrightarrow{BP}$ has length

$$|BP|=|AB|\cos\beta=|v|\cos\beta,$$

and is obtained from $v$ by rotation through $-\beta$ because the triangle is constructed externally. Hence

$$\overrightarrow{BP}=\cos\beta,R_{-\beta}v.$$

Similarly, from the right triangle $BQC$,

$$\overrightarrow{BQ}=\cos\beta,R_{\beta}w.$$

Let $K$ be the midpoint of $AC$. Since

$$A=B+v,\qquad C=B+w,$$

we have

$$K=B+\frac{v+w}{2}.$$

Therefore

$$\overrightarrow{KP} = \overrightarrow{BP}-\overrightarrow{BK} = \cos\beta,R_{-\beta}v-\frac{v+w}{2},$$

and

$$\overrightarrow{KQ} = \cos\beta,R_{\beta}w-\frac{v+w}{2}.$$

Using

$$\cos\beta,R_{-\beta} = \begin{pmatrix} \cos^2\beta & \sin\beta\cos\beta\ -\sin\beta\cos\beta & \cos^2\beta \end{pmatrix},$$

a direct calculation gives

$$\cos\beta,R_{-\beta}-\frac12I = -\frac12R_{-2\beta}.$$

Hence

$$\overrightarrow{KP} = -\frac12R_{-2\beta}(w-v).$$

Since

$$\cos\beta,R_{\beta}-\frac12I = \frac12R_{2\beta},$$

we obtain

$$\overrightarrow{KQ} = \frac12R_{2\beta}(w-v).$$

Because rotations preserve lengths,

$$KP=\frac12|w-v|,\qquad KQ=\frac12|w-v|.$$

Thus

$$KP=KQ.$$

Consequently, triangle $PQK$ is isosceles.

Now compare the directions of $\overrightarrow{KP}$ and $\overrightarrow{KQ}$:

$$\overrightarrow{KP} = -\frac12R_{-2\beta}(w-v), \qquad \overrightarrow{KQ} = \frac12R_{2\beta}(w-v).$$

The vector $-!R_{-2\beta}(w-v)$ is obtained from $w-v$ by a rotation through

$$180^\circ-2\beta,$$

while $R_{2\beta}(w-v)$ is obtained from $w-v$ by a rotation through

$$2\beta.$$

Hence the angle between $\overrightarrow{KP}$ and $\overrightarrow{KQ}$ equals

$$(180^\circ-2\beta)-2\beta = 180^\circ-4\beta.$$

Therefore

$$\angle PKQ=180^\circ-4\beta.$$

Since triangle $PQK$ is isosceles with $KP=KQ$,

$$\angle KPQ=\angle KQP = \frac{180^\circ-(180^\circ-4\beta)}2 = 2\beta.$$

Thus the angles of triangle $PQK$ are

$$\boxed{\angle KPQ=\angle KQP=2\beta,\qquad \angle PKQ=180^\circ-4\beta.}$$

Verification of Key Steps

The first delicate step is the formula

$$\overrightarrow{BP}=\cos\beta,R_{-\beta}v.$$

In triangle $APB$, the hypotenuse is $AB$, so

$$|BP|=|AB|\cos\beta.$$

The direction of $BP$ is obtained by turning $BA$ toward the exterior of triangle $ABC$. Reversing the sign of the rotation would place $P$ on the wrong side of $AB$ and would change the final answer.

The second delicate step is the identity

$$\cos\beta,R_{-\beta}-\frac12I = -\frac12R_{-2\beta}.$$

Writing both sides as matrices,

$$\cos\beta,R_{-\beta}-\frac12I = \begin{pmatrix} \cos^2\beta-\frac12 & \sin\beta\cos\beta\ -\sin\beta\cos\beta & \cos^2\beta-\frac12 \end{pmatrix},$$

and

$$-\frac12R_{-2\beta} = \begin{pmatrix} -\frac12\cos2\beta & \frac12\sin2\beta\ -\frac12\sin2\beta & -\frac12\cos2\beta \end{pmatrix}.$$

Using

$$\cos^2\beta-\frac12=\frac12\cos2\beta-\frac12=-\frac12(1-\cos2\beta) =-\frac12\cos2\beta,$$

together with

$$\sin2\beta=2\sin\beta\cos\beta,$$

the matrices coincide.

The third delicate step is the determination of the angle at $K$. The vectors are rotated copies of the same vector:

$$\overrightarrow{KP}=-R_{-2\beta}u,\qquad \overrightarrow{KQ}=R_{2\beta}u.$$

The minus sign contributes exactly $180^\circ$. Omitting it would yield $4\beta$ instead of $180^\circ-4\beta$, which is incorrect because the vertex angle of the isosceles triangle must be supplementary to the sum of the equal base angles.

Alternative Approaches

A synthetic solution can be obtained by expressing the positions of $P$ and $Q$ through spiral similarities. Since

$$\frac{BP}{BA}=\frac{BQ}{BC}=\cos\beta,$$

and the corresponding rays differ from $BA$ and $BC$ by equal angles, there are rotations and homotheties centered at $B$ carrying $A$ to $P$ and $C$ to $Q$. The midpoint $K$ of $AC$ then transforms into the midpoint of a suitable rotated image of $AC$. Chasing the resulting rotations shows that $KP$ and $KQ$ are equal and differ in direction by $180^\circ-4\beta$.

The vector-rotation method is preferable because it converts the geometric construction directly into two compact formulas. Once the expressions for $\overrightarrow{KP}$ and $\overrightarrow{KQ}$ are obtained, both the equality of the sides and the values of the angles follow immediately from properties of rotations.