A circular race track has three stop points A, B and C on it. Three straight tracks AB, AC and BC connect these points to each other. The equations of the straight line AC and BC are given by: y = 2x + 6, and y = (-0.5)x + 8 respectively. The length of the paths BC is 80 meters while the length of AB is (5/4) the length of BC. The radius of the race track is:

A. 140 meters B. 100 meters C. 70 meters D. 50 meters E. It can’t be determined.

Let us analyze solution of this problem step by step.

What math objects do we have? We have a circle with three points on it joined by line segments, making a triangle ABC inscribed in a circle. We know equations of the lines BC and AC. We also know the lengths of line segments BC and AB. (AB = 5/4 × 80 = 100).

What do we have to find? We need to find radius of the circle.

Well, it seems we have a lot of information but no clear way of calculating a radius from it. So let us draw a rough sketch to visualize it:

Looking at the sketch let us think of approach we can use to find radius.

The first approach I am thinking of is the most general one. I'm applying methods that would work in any case. Here it is:

Since we know nothing about circle itself we need to use inscribed triangle ABC. There is a formula for calculating a radius of circumscribed circle but we need to know the lengths of all sides. How can we calculate the length of AC?

In order to calculate the length of AC we need to know the coordinates of points A and C. Since C is the cross point of lines AC and BC we can calculate the coordinates of C by solving system of two line equations:

y = 2x + 6 y = (-0.5)x + 8

Then we can make a system of two equations finding coordinates of B.

B lies on a line BC, so one equation is: y = (-0.5)x + 8. B is 80 m away from C. So another equation is the one for distance between B and C.

Solving this system we will get two choices for coordinates of point B, because there are two possible points on a line BC that are 80m away from C (one on each side). Another option is marked in blue on this image:

We can pick any one of those because we assume data given in the question is sufficient so in either case radius should be the same.

Then we can make a system of two equations finding coordinates of A.

A lies on a line AC, so one equation is: y = 2x + 6. A is 80 m away from B. So another equation is the one for distance between A and B.

Solving this system we will also get two choices for coordinates of point A, because there are two possible points on a line BC that are 100m away from B (another option is marked in green).

We can pick any one of those because we assume data given in the question is sufficient so in either case radius should be the same.

So far we would find coordinates of points A, B and C. So we can find length of AC and use formula for radius of circumscribed circle.

That is quite a long way to go and we even didn't actually solve any of the systems of equations. It will definitively lead us to finding the answer but is there another way?

In this case there is. We can see that out triangle is not so arbitrary as it seems. If we take a look at the equations of lines AC and BC we can see that they are perpendicular because the product of the slopes is (-0.5) × 2 = -1. Therefore angle BCA is 90 degrees.

If a right triangle is inscribed in a circle then its hypotenuse is the diameter of that circle. The radius is half of that. So radius is AB / 2 = 100 / 2 = 50.

The answer is 50.

What can I say about this "shortcut" in solving? Practice, be concentrated and be attentive to all the details and you will find many more "shortscuts"!

Users browsing this forum: No registered users and 0 guests

You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum

GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.