# Puzzle relation with geomtry

An army $$40$$ km long is moving at constant speed. A horseman takes a message from the rear end of the army to the general in the front end of the army and returns back to the rear end of the army (moving with constant speed). During this time, the army moves a total of $$40$$ km.

What total distance did the horseman travel?

To me, we can draw place-time $$(x,t)$$-plane.

HINT

The sketch might help if velocity $$V$$ is taken properly. Note the horseman's track. (Sketch would be moved to answers area later.)

• Can you edit your question to add your thoughts and ideas about it? – 5xum Nov 14 '18 at 9:14
• @5xum, I edited. – 1ENİGMA1 Nov 14 '18 at 9:32
• @1ENİGMA1 There is also a nice geometric interpretation, I'll post that later. – user Nov 14 '18 at 15:05

## 2 Answers

Let indicate with $$v_a$$ the speed of the army, by $$v_h$$ the speed of the horseman, then for the travel in front we have

• the position of the general is: $$x_g=40+v_at$$

• the position of the horseman is: $$x_h=v_ht$$

and

$$x_g=x_h\implies t_1=\frac{40}{v_h-v_a}$$

For the travel to return we have

• the position of the rear: $$x_r=40-v_at$$

• the position of the horseman is: $$x_h=v_ht$$

and

$$x_r=x_h\implies t_2=\frac{40}{v_h+v_a}$$

From the given condition we know that

$$(t_1+t_2)v_a=40$$

that is

$$40\frac{v_a}{v_h-v_a}+40\frac{v_a}{v_h+v_a}=40$$

$$\frac{2v_av_h}{v_h^2-v_a^2}=1 \implies v_h^2-2v_av_h-v_a^2=0 \implies v_h=(1+\sqrt2)v_a$$

and the distance traveled is therefore $$40(1+\sqrt2)$$ km.

With reference to the followig picture assuming wlog $$Q=(1,1)$$ the coordinates of the point $$P$$ are

$$P=(0,1)+t(1,1) \quad t\in[0,1]$$

and from the given condition of constant speed by the horseman we obtain

$$\frac{1+t}{t}=\frac{1+t-1}{1-t} \implies 1-t^2=t^2 \implies t=\frac{\sqrt2}2$$

therefore $$P=\left(\frac{\sqrt2}2,1+\frac{\sqrt2}2\right)$$

and the distance travelled is equal to

$$\frac d L=1+\frac{\sqrt2}2+\frac{\sqrt2}2=1+\sqrt 2$$