Kvant Math Problem 661

Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m40s
Source on kvant.digital

Problem

On the shore of a circular lake there are four piers $K$, $L$, $P$, $Q$. A motorboat departs from pier $K$, and a rowboat from $L$. If the motorboat heads straight for $P$ and the rowboat straight for $Q$, they will collide at some point $X$ in the lake. Prove that if the motorboat heads for $Q$ and the rowboat for $P$, then they will reach these piers simultaneously.

N. B. Vasilyev

Exploration

Let the speeds of the motorboat and rowboat be constant, equal to $v_M$ and $v_R$. In the first situation the boats travel along the chords $KP$ and $LQ$ and collide at a point $X$.

Since they collide, the travel times from departure to $X$ are equal. Hence

$$\frac{KX}{v_M}=\frac{LX}{v_R}.$$

This gives

$$\frac{v_M}{v_R}=\frac{KX}{LX}.$$

To prove the statement, when the destinations are exchanged we must show that the travel times to $Q$ and $P$ satisfy

$$\frac{KQ}{v_M}=\frac{LP}{v_R},$$

or equivalently

$$\frac{KQ}{LP}=\frac{v_M}{v_R}.$$

Thus everything reduces to proving

$$\frac{KQ}{LP}=\frac{KX}{LX}.$$

The point $X$ is the intersection of the chords $KP$ and $LQ$ of the same circle. The classical intersecting chords theorem gives

$$KX\cdot XP=LX\cdot XQ.$$

This yields

$$\frac{KX}{LX}=\frac{XQ}{XP}.$$

The desired relation would follow if

$$\frac{XQ}{XP}=\frac{KQ}{LP}.$$

At this stage one should look for similar triangles. Since $X$ lies on both chords, the angle between $XQ$ and $XP$ equals the angle between $LQ$ and $KP$. In the cyclic quadrilateral $KLPQ$,

$$\angle KQL=\angle KPL$$

because both subtend the arc $KL$. Therefore

$$\angle XQP=\angle KLP.$$

Similarly,

$$\angle XPQ=\angle KQL.$$

Hence triangles $XQP$ and $KLP$ are similar, giving

$$\frac{XQ}{XP}=\frac{LP}{KL} \Big/ \frac{KP}{KL} =\frac{LP}{KP},$$

which is not the desired expression. This indicates that the correspondence must be chosen carefully.

Checking the angles again,

$$\angle XQP=\angle LQP, \qquad \angle KLP=\angle KLP.$$

Both subtend arc $LP$, so these are equal. Likewise,

$$\angle XPQ=\angle KPQ,$$

and $\angle KLQ$ subtends the same arc $KQ$. Thus triangles $XQP$ and $XLK$ are actually similar. Then

$$\frac{XQ}{XP}=\frac{LK}{KX}.$$

Combining with the chord theorem gives nothing immediate.

A better route is to use the same similarity more systematically. From the cyclic quadrilateral,

$$\angle XQP=\angle KLP, \qquad \angle XPQ=\angle KLQ.$$

Since $\angle KLP=\angle KQP$ and $\angle KLQ=\angle KPQ$, triangles $XQP$ and $LPK$ are similar. Then

$$\frac{XQ}{XP}=\frac{LP}{PK}.$$

Together with

$$\frac{KX}{LX}=\frac{XQ}{XP},$$

we obtain

$$\frac{KX}{LX}=\frac{LP}{PK}.$$

This still does not match the target, so this avenue is not the correct one.

The crucial point is to identify the right pair of similar triangles. Consider triangles $KXQ$ and $LXP$. They have

$$\angle KXQ=\angle LXP$$

as vertical angles. Also,

$$\angle KQX=\angle KQL, \qquad \angle LPX=\angle LPK,$$

and these are equal because both subtend arc $KL$. Therefore the triangles are similar. This immediately yields

$$\frac{KX}{LX}=\frac{KQ}{LP},$$

which is exactly the relation needed.

The potentially dangerous step is the similarity of triangles $KXQ$ and $LXP$; every angle correspondence must be checked carefully using the cyclicity of the four piers.

Problem Understanding

The four piers $K,L,P,Q$ lie on the circumference of a circular lake. A motorboat leaves $K$ and a rowboat leaves $L$. When the motorboat heads directly toward $P$ and the rowboat directly toward $Q$, they collide at a point $X$ inside the lake.

The collision implies that the time taken by the motorboat to travel from $K$ to $X$ equals the time taken by the rowboat to travel from $L$ to $X$. The task is to prove that if the destinations are exchanged, so that the motorboat travels from $K$ to $Q$ and the rowboat from $L$ to $P$, then the two boats reach their destinations at the same time.

This is a Type B problem. The core difficulty is to convert the equality of times in the first voyage into a geometric relation among the chords of the circle, and then show that this relation implies equality of times in the second voyage.

Proof Architecture

The first claim is that the collision condition implies

$$\frac{v_M}{v_R}=\frac{KX}{LX},$$

where $v_M$ and $v_R$ are the constant speeds of the motorboat and rowboat. This follows directly from equality of travel times to the collision point.

The second claim is that triangles $KXQ$ and $LXP$ are similar. They have one pair of vertical angles at $X$, and a second pair of equal inscribed angles subtending the same arc $KL$.

The third claim is that the similarity yields

$$\frac{KX}{LX}=\frac{KQ}{LP}.$$

This comes from corresponding sides of the similar triangles.

Combining the first and third claims gives

$$\frac{v_M}{v_R}=\frac{KQ}{LP},$$

which is equivalent to equality of travel times in the second voyage.

The most delicate lemma is the similarity of triangles $KXQ$ and $LXP$, because the correct correspondence of angles must be established from the cyclic quadrilateral.

Solution

Let the constant speeds of the motorboat and rowboat be $v_M$ and $v_R$.

In the first situation the motorboat travels along the chord $KP$ and the rowboat along the chord $LQ$. They collide at the point $X$, which is the intersection of these two chords.

Since they start simultaneously and collide at $X$, the travel times to $X$ are equal. Hence

$$\frac{KX}{v_M}=\frac{LX}{v_R},$$

so

$$\frac{v_M}{v_R}=\frac{KX}{LX}.$$

It remains to relate the ratio $KX/LX$ to the lengths $KQ$ and $LP$.

Consider triangles $KXQ$ and $LXP$.

Because the chords $KP$ and $LQ$ intersect at $X$,

$$\angle KXQ=\angle LXP,$$

these being vertical angles.

Next,

$$\angle KQX=\angle KQL,$$

since $Q,X,L$ are collinear, and

$$\angle LPX=\angle LPK,$$

since $P,X,K$ are collinear.

The quadrilateral $KLPQ$ is cyclic. The inscribed angles $\angle KQL$ and $\angle LPK$ subtend the same arc $KL$, therefore

$$\angle KQX=\angle LPX.$$

Thus triangles $KXQ$ and $LXP$ are similar.

Corresponding sides of these triangles satisfy

$$\frac{KX}{LX}=\frac{KQ}{LP}.$$

Combining this with the relation obtained from the collision condition gives

$$\frac{v_M}{v_R} = \frac{KX}{LX} = \frac{KQ}{LP}.$$

Rearranging,

$$\frac{KQ}{v_M} = \frac{LP}{v_R}.$$

The left-hand side is the time required for the motorboat to travel from $K$ to $Q$, and the right-hand side is the time required for the rowboat to travel from $L$ to $P$.

Hence, when the motorboat heads for $Q$ and the rowboat heads for $P$, they reach those piers simultaneously.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the conversion of the collision condition into a speed ratio. The collision means that both boats have traveled for exactly the same amount of time. If that common time is $t$, then $KX=v_M t$ and $LX=v_R t$. Dividing gives

$$\frac{v_M}{v_R}=\frac{KX}{LX}.$$

No additional assumptions are involved.

The second delicate step is the similarity of triangles $KXQ$ and $LXP$. The equality

$$\angle KXQ=\angle LXP$$

comes from intersecting lines $KP$ and $LQ$. The equality

$$\angle KQX=\angle LPX$$

requires two substitutions: $Q,X,L$ are collinear and $P,X,K$ are collinear, after which the angles become $\angle KQL$ and $\angle LPK$. These are equal because both subtend the arc $KL$ of the circumcircle. This establishes similarity by the angle-angle criterion.

The final step is extracting the correct ratio from the similarity. Under the correspondence

$$KX \leftrightarrow LX,\qquad KQ \leftrightarrow LP,$$

the similarity gives

$$\frac{KX}{LX}=\frac{KQ}{LP}.$$

Any incorrect matching of vertices would lead to a different ratio and would not yield the desired conclusion.

Alternative Approaches

One may use the intersecting-chords theorem together with trigonometric forms of the law of sines. From the cyclic quadrilateral, the ratios of segments cut by the intersecting chords can be expressed through equal subtended arcs. After simplifying, one again obtains

$$\frac{KX}{LX}=\frac{KQ}{LP}.$$

The proof above is preferable because it avoids auxiliary computations. Once the correct pair of triangles, $KXQ$ and $LXP$, is identified, the desired relation follows from a single similarity argument and the conclusion becomes immediate from the equality of travel times to the collision point.