Kvant Math Problem 17

Let the fork be at point $A$.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m43s
Source on kvant.digital

Problem

A peasant, having come to a fork where two roads diverged at an angle of $60\degree$, asked, “How do I get to the village of $NN$?” He was told: “Take the left road to the village of $N$, which is eight versts from here. There you will see a broad straight road branching off to the right at a right angle; that is exactly the road to $NN$. Or you can go another way: take the right road now; when you reach the railway, that means you have covered half the journey. Then turn left and walk straight along the tracks all the way to $NN$.” “Well, which route is shorter?” “It makes no difference; one way is just as long as the other.” And the peasant set off along the right road. How many versts will he have to walk to reach $NN$? More than ten or less? And what if he goes directly from the fork to $NN$? (All roads are assumed to be straight.)

Exploration

Let the fork be at point $A$. The left road goes from $A$ to the village $N$, call that point $B$, with $AB=8$. The angle between the two roads is $60^\circ$.

From $B$ a road to $NN$ branches off at a right angle to the right. Since the traveler approached $B$ along $AB$, the road $BQ$ to $NN$ is perpendicular to $AB$. Let $Q$ denote the village $NN$.

The second route starts along the right road from $A$. Let the railway meet that road at a point $P$. The instructions say that reaching the railway means half the journey has been completed. Since after reaching $P$ one walks along the railway to $Q$, the total length of this route is

$$AP+PQ,$$

and $AP$ is half of it. Hence

$$AP=\frac{AP+PQ}{2},$$

so

$$PQ=AP.$$

Thus $P$ lies on the right road from $A$, and the railway segment from $P$ to $Q$ has the same length as $AP$. Since the railway is reached and then followed after turning left, the railway must be perpendicular to the right road. Hence triangle $APQ$ is right isosceles at $P$.

Choose coordinates. Put

$$A=(0,0),\qquad B=(8,0).$$

Take the right road to make an angle $60^\circ$ below the positive $x$-axis. Then points on it have form

$$P=\left(\frac t2,-\frac{\sqrt3,t}{2}\right),$$

where $t=AP$.

Since triangle $APQ$ is right isosceles and turning left from the road means rotating the direction of $AP$ by $90^\circ$ counterclockwise, the vector $PQ$ equals

$$\left(\frac{\sqrt3,t}{2},\frac t2\right).$$

Hence

$$Q=\left(\frac{1+\sqrt3}{2}t,\frac{1-\sqrt3}{2}t\right).$$

But $Q$ lies on the road from $B$ perpendicular to $AB$, namely the vertical line $x=8$. Therefore

$$\frac{1+\sqrt3}{2}t=8,$$

giving

$$t=\frac{16}{1+\sqrt3}=8(\sqrt3-1).$$

The length of the second route is

$$2t=16(\sqrt3-1).$$

Numerically this is about $11.71$, so it exceeds $10$.

The direct distance is

$$AQ=\sqrt{2},t=8\sqrt2,(\sqrt3-1).$$

Numerically this is about $8.28$.

The only delicate point is the interpretation of “half the journey”. It refers to half of the length of the chosen route, which yields $PQ=AP$ and produces the right isosceles triangle.

Problem Understanding

We are given two straight roads starting from a fork at an angle of $60^\circ$. The left road reaches village $N$ after $8$ versts, and from $N$ a road perpendicular to it leads to village $NN$. Alternatively, one may take the right road to a railway, and reaching the railway means exactly half of the total route has been completed; after that one follows the railway straight to $NN$. The two routes have equal length.

This is a Type C problem. We must determine the length of the route taken along the right road and railway, decide whether it exceeds $10$ versts, and find the direct distance from the fork to $NN$.

The answer should be

$$16(\sqrt3-1)\ \text{versts}$$

for the route length, which is greater than $10$, and

$$8\sqrt2,(\sqrt3-1)\ \text{versts}$$

for the direct distance.

The core difficulty is translating the statement that the railway is reached after half the journey into a geometric condition.

Proof Architecture

The first lemma is that if reaching the railway means half the journey has been completed, then the railway segment from the meeting point to $NN$ has the same length as the initial road segment. This follows from a simple equation for the total route length.

The second lemma is that the second route forms a right isosceles triangle. The railway is perpendicular to the right road because the traveler turns left from that road and then proceeds straight along the tracks; combined with the first lemma this yields a right triangle with equal legs.

The third lemma is that the coordinates of $NN$ can be expressed in terms of the common leg length $t$. This follows from elementary coordinate geometry.

The fourth lemma is that the coordinate condition expressing that $NN$ lies on the road from $N$ determines $t$ uniquely. Solving the resulting linear equation gives the route length.

The most delicate step is the conversion of the phrase “half the journey” into the equality of the two legs of the right triangle.

Solution

Let $A$ be the fork, let $B$ be the village $N$, and let $Q$ be the village $NN$.

Choose coordinates so that

$$A=(0,0),\qquad B=(8,0).$$

The left road is the positive $x$-axis. Since the two roads form an angle of $60^\circ$, take the right road to be the ray making an angle $60^\circ$ below the positive $x$-axis.

Let $P$ be the point where the right road meets the railway.

Denote

$$AP=t.$$

The second route has length

$$AP+PQ=t+PQ.$$

According to the statement, reaching the railway means that half the journey has been completed. Hence

$$t=\frac{t+PQ}{2},$$

which implies

$$PQ=t.$$

At $P$ the traveler turns left from the right road and then proceeds along the railway. Consequently the railway is perpendicular to the right road. Thus triangle $APQ$ is right-angled at $P$, and since $AP=PQ=t$, it is a right isosceles triangle.

The direction vector of the right road is

$$\left(\frac12,-\frac{\sqrt3}{2}\right).$$

Therefore

$$P=\left(\frac t2,-\frac{\sqrt3,t}{2}\right).$$

A left turn through $90^\circ$ changes this direction vector into

$$\left(\frac{\sqrt3}{2},\frac12\right).$$

Since $PQ=t$,

$$\overrightarrow{PQ} = \left(\frac{\sqrt3,t}{2},\frac t2\right).$$

Hence

$$Q = \left(\frac t2,-\frac{\sqrt3,t}{2}\right) + \left(\frac{\sqrt3,t}{2},\frac t2\right) = \left(\frac{1+\sqrt3}{2}t,\frac{1-\sqrt3}{2}t\right).$$

The road from $B$ to $Q$ is perpendicular to $AB$. Since $AB$ is horizontal, this road is the vertical line

$$x=8.$$

Because $Q$ lies on it,

$$\frac{1+\sqrt3}{2}t=8.$$

Thus

$$t=\frac{16}{1+\sqrt3} =8(\sqrt3-1).$$

The length of the second route equals

$$AP+PQ=2t=16(\sqrt3-1).$$

To determine whether this exceeds $10$, compute

$$16(\sqrt3-1)\approx16(0.732)=11.712,$$

so the route is longer than $10$ versts.

Finally, triangle $APQ$ is right isosceles with legs $t$, hence

$$AQ=\sqrt{t^2+t^2}=t\sqrt2.$$

Substituting the value of $t$ gives

$$AQ=8\sqrt2,(\sqrt3-1).$$

Since

$$8\sqrt2,(\sqrt3-1)\approx8.28,$$

the direct road from the fork to $NN$ would be about $8.28$ versts long.

Therefore the route actually taken has length

$$\boxed{16(\sqrt3-1)\ \text{versts}},$$

which is greater than $10$, and the direct distance from the fork to $NN$ is

$$\boxed{8\sqrt2,(\sqrt3-1)\ \text{versts}}.$$

Verification of Key Steps

The statement that reaching the railway means half the journey has been completed must be interpreted quantitatively. If the total route length is $L$, then $AP=L/2$. Since $L=AP+PQ$, one obtains

$$AP=\frac{AP+PQ}{2},$$

hence

$$PQ=AP.$$

Any other interpretation would contradict the wording that half of the journey has already been covered upon arriving at the railway.

The coordinate formula for $Q$ can be checked independently. The vector from $A$ to $P$ has length $t$ and direction $-60^\circ$, namely

$$\left(\frac t2,-\frac{\sqrt3,t}{2}\right).$$

The vector from $P$ to $Q$ has the same length and direction $30^\circ$, namely

$$\left(\frac{\sqrt3,t}{2},\frac t2\right).$$

Adding them yields exactly the coordinates used in the proof.

The final numerical comparison should also be checked directly. Since

$$\sqrt3\approx1.732,$$

we have

$$16(\sqrt3-1)\approx11.712>10.$$

A mistaken sign in the rationalization step would instead give $8(1-\sqrt3)$, a negative length, revealing the error immediately.

Alternative Approaches

A purely synthetic solution avoids coordinates. Let $AP=PQ=t$. Since the angle between the roads at $A$ is $60^\circ$ and the railway is perpendicular to the right road, the angle between $AQ$ and the road to the railway equals $45^\circ$. Constructing the right isosceles triangle $APQ$ and projecting onto the direction of $AB$ shows that the horizontal displacement from $A$ to $Q$ equals

$$t\cos60^\circ+t\cos30^\circ = \frac{1+\sqrt3}{2}t.$$

Since this displacement equals $AB=8$, the same equation

$$\frac{1+\sqrt3}{2}t=8$$

is obtained.

The coordinate method is preferable because every geometric condition translates directly into a simple equation, and the determination of both route lengths becomes completely straightforward.