Running, Running, Running!

Because the girls together run half of the track between the start and the first meeting and the whole track between the first and second meetings and they are running at constant speeds, the time between meetings must be twice that between the start and the first meeting. The first girl then runs 200 meters between the two meetings. Hence, between the two meetings, the girls run $200+150=\boxed{350}$ meters: one track length.

Let the track have length $L.$ Since the girls start at opposite points on the track, they run a combined distance of $\dfrac L2$ meters (half the length of the track) to get to their first meeting point, so the first girl runs $100$ meters and the second girl runs $\dfrac L2 - 100$ meters. Then, to get to their second meeting point, they run a combined distance of $L$ meters (the whole track's length), so the first girl runs $L - 150$ meters while the second girl runs $150$ meters.

Because the two girls run at constant speeds, the ratios of their distances traveled must be equal. (For example, if the second girl is always twice as fast as the first girl, then both between the start & the first meeting point and between the first and second meeting points, the second girl must run twice as far as the first girl.) We must have $\dfrac{100}{\dfrac L2 - 100} = \dfrac{L-150}{150}.$

Cross-multiplying, we have $15000 = \dfrac{L^2}{2} - 175L + 15000,$ or $0 = \dfrac{L^2}{2} - 175L.$ Multiplying by $2$ gives $0 = L^2 - 350L,$ which factors as $0 = L(L-350).$ Therefore, since clearly $L \neq 0,$ the length of the track must be $L = \boxed{350} \text{ m}.$

Instead of bashing this with algebra, let's think about it.

When they ran the first time, they ran a total of $\frac{1}{2}$ of the track and the first girl ran $100$ m. When they met the second time, they ran the length of the entire track and the second girl ran $150$ m of it and the first girl must have ran $100\cdot2=200$ m of it (since this length is twice the length of the previous length so she would've ran twice the distance). Hence the length of the track is $150+200=\boxed{350}$ m.

Let the radius of the circular track be $r$ . Let the speed of the first girl be $v_1$ and the speed of the second be $v_2$ .

Since they both started running towards each other at the same time, initially separated at a distance of $\pi r$ , the time that they took to run to the first meeting point is the same.

The first girl ran $100m$ while the second girl ran $(\pi r - 100)m$ .

$\Longrightarrow t_1 = \dfrac{100}{v_1} = \dfrac{\pi r - 100}{v_2}$

$\Longrightarrow \dfrac{v_1}{v_2} = \dfrac{100}{\pi r - 100}$

Now, let us consider the second meeting. Here, the second girl ran $150m$ while the first ran $(2 \pi r - 150)m$ in the same time, still at the same speeds.

$\Longrightarrow t_2 = \dfrac{150}{v_2} = \dfrac{2 \pi r -150}{v_1}$

$\Longrightarrow \dfrac{v_1}{v_2} = \dfrac{2 \pi r - 150}{150}$

Equating the ratios of their speeds ( $\frac{v_1}{v_2}$ ):

$\Longrightarrow \dfrac{100}{\pi r - 100} = \dfrac{2 \pi r - 150}{150}$

$\Longrightarrow 2 \pi^{2} r^{2} - 350 \pi r = 0$

$\Longrightarrow 2\pi r = \boxed{350}$

Main idea : The two girls combined trace a half-circle before the first meeting and then a full circle between the first meeting and the second meeting.

Complete proof :

Call the two girls $A, B$ respectively. Change the frame of reference to stick with the first runner. This way, the first runner is considered stationary and the second runner has their combined speed; call this second girl $C$ .

C runs half a circle to meet with the first runner, and then another full circle to meet again the second time. Let the times be $t_0, t_1, t_2$ for the initial run, the first meeting, and the second meeting respectively. Since C takes twice the distance of the first meeting to meet the second time, it follows that $t_2 - t_1 = 2(t_1 - t_0)$ . Also, WLOG assume that $t_0 = 0, t_1 = 1$ , so $t_2 = 3$ .

A runs $100$ meters per unit of time (between $t_0$ and $t_1$ ). B runs $150$ meters for $2$ units of time (between $t_1$ and $t_2$ ), or $75$ meters per unit of time. Combined, their total speed is $175$ meters per unit of time, which is also the speed of C. Since C takes $2$ units of time to circle the track, it follows that the length of the track is $175 \cdot 2 = \boxed{350}$ meters.

Motivation : To be fair I've seen this kind of problems before, so I know the method that doesn't involve messy algebra. I didn't change the frame of reference when solving, and instead think "the two girls combined trace this much distance before the first meeting" and so, but it's equally messy to write for this solution, so why not.

Generalization : As the problem is worded now, if the first girl runs for $x$ meters before the first meeting and the second girl runs for $y$ meters after the first meeting but before the second meeting, then the answer is $2x+y$ meters. By manipulating other things (they run in the same direction, they start at a different angle than $\pi$ radians from the center, there are more than two girls), there can be much more variations, which mostly follow the same basic idea.

First time they run half of the circumference. Second time they run full circumference.

First time first runs 100 meters, thus second time she runs 2*100 = 200 meters.

Since second time (when they run full cimcufrence) Brenda runs 200 meters and secound runs 150 meters, thus the circumference is 200 + 150 = 350 meters

2(50+75)

1st meeting point: Velocity of girl 1 =100/t, Velocity of girl 2=Pi r/t 2nd meeting point: Velocity of girl 1 =(2Pi r-150)/t', Velocity of girl 2=150/t' taking ratios of velocities and equating them for 1st meeting point and second meeting point, we get the value of Pi*r=175 or circumference=350!

100+(100+150)=350

First, we note that the ratio of the distances traveled is constant. Thus we get that $\frac{l/2 - 100}{100}=\frac{150}{l-150}$ $30000 = (l/2-100)(l-150) = l^{2}-350+30000$ $l^{2}-350 l=0$ $l=\boxed{350}$

Let the track length be L, speed of the girls be x1 and x2. Equating time for the first meeting, 100/x1=(L/2-100)/x2 Equating time for the second meeting, (L-150)/x1=150/x2 Eliminating x1 and x2, solve for L. L^2 - 350L = 0 L=350

Let u and v be the velocities of two girls. t be the time after which they meet for first time After first meet ut = 100 πr =ut +vt πr = 100 +vt .........1

Now lets look at the situation from 1st point of meeting. Now the time taken by them to cover half of perimeter is t. Then time taken by them to cover the whole cirlce is 2t The distance moved by both girls when they meet at 2nd point(from first point) is perimeter of circle we have v(2t) = 150...................2

solving 1 and 2, vt = 75 there fore πr = 100+75 2πr = 350

Let Track Length = L, v1 = Speed of first girl, v2 = Speed of second girl and t=time till first meeting. 100 = v1 t and (L/2 - 100) =v2 t or v2 =(L/2 - 100) v1/100. From the second meeting, 150 100/(v1)(L/2-100) =(L-150)/v1, Solve to get, L= 0 or 350 met; hence the latter.

Call the length of the race track x. When they meet at the first meeting point, the first girl has run 100 meters, while the second girl has run (x/2)-100 meters. By the second meeting point, she has run 150 meters, while the first girl has run x - 150 meters. Since they run at a constant speed, we can set up a proportion: (100/(x-150))=(x/2-100)/150. Cross-multiplying, we get that x=350m

First, let the first girl be A and second girl be B and the circumference of the track be c. In the first time interval, girl A ran 100 meters while girl B ran half of the track minus 100 meters (c/2-100). In the second time interval, girl A ran the whole track minus 150 meters (c-150), while girl B ran 150 meters.

Since the two girls run at constant speed, we can set up equivalent proportion of the distance they traveled during the same amount of time. These are the ratio we have got with B:A,

c/2-100: 100

150: c-150

Next, set up a proportion with B over A:

(c/2-100)/100 = 150/(c-150)

After simplifying, we get a quadratic equation:

c^{2} - 350c = 0

c(c-350)=0

c=350

Therefore, the track is 350 meters.