Eight points are equally spaced on a circle. If 3 of the 8 points are

Question

Eight points are equally spaced on a circle. If 3 of the 8 points are to be selected at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?

A) 1/14
B) 1/7
C) 3/14
D) 3/7
E) 6/7

nick1816 · Accepted Answer

Total number of triangle possible= 8C3=56
Number of ways to select the 2 points that are making the diameter=4
Number of triangles can me made using 2 points that are making a diameter=6 (We can select any of the remaining 6 points)
Total number of right triangles possible=6*4=24

probability that a triangle having the 3 points chosen as vertices will be a right triangle= 24/56=3/7

BrentGMATPrepNow · Answer

TOUGH question!!! Key property: An inscribed angle that contains (aka "holds") the DIAMETER will be a 90-degree angle. For example, let's draw one of the diameters... Eight points are equally spaced on a circle. If 3 of the 8 points1.png The inscribed angle that contains (aka "holds") the DIAMETER will be a 90-degree angle. Eight points are equally spaced on a circle. If 3 of the 8 points2.png Likewise, this inscribed angle also contains (aka "holds") the DIAMETER, so it will also be a 90-degree angle. Eight points are equally spaced on a circle. If 3 of the 8 points3.png So, for the ONE PARTICULAR diameter (shown below)... Eight points are equally spaced on a circle. If 3 of the 8 points4.png ...we can see that, if we make any of the 6 points the 3rd vertex of the triangle, we will get a right triangle. This means that, FOR EACH diameter in our circle, there are 6 points that will create a right triangle. Since there are 4 diagonals altogether.... Eight points are equally spaced on a circle. If 3 of the 8 points5.png ....we know that the TOTAL number of right triangles possible = ( 4 )( 6 ) = 24 --------------------------------------- Now we need to determine how many different triangles can be created by selecting 3 of the 8 points. Since the order in which we select the 3 points does not matter, we can use COMBINATIONS. We can select 3 points from 8 points in 8C3 ways 8C3 = (8)(7)(6)/(3)(2)(1) = 56 ----------------------------------------- So, P(we get a right triangle) = ( total number of RIGHT triangles possible )/( total number of triangles possible ) = 24 / 56 = 3/7 Answer: D RELATED VIDEO https://www.youtube.com/watch?v=xiyv2MacIUk

Lampard42 · Answer

Total number of triangles will be 8c3 = 8*7

Total number of triangles with right angles would be where diameter of the circle is one sides of the triangle. So 24 such triangles are possible.

Probability is 3*8/(8*7)=3/7

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Archit3110 · Answer

given points are equally spaced so mark points opposite to each other
total ∆ possible 8c3 = 56
for right angled first point can be chosen in 8c1 ways
2nd point in 3c1 & 3rd point 1c1 ; 8c1*3c1*1c1 = 24
so P of right ∆ 24/56 ; 3/7 
IMO D

LeoN88 · Answer

I think the highlighted part will be= (1/2) * 8C1* 6C1 * 1C1= 48/2= 24

why 6C1? because you have to consider both the sides of the semicircle, when you choose one point the semi circle can be formed at either point.
why divide by half? Let AB be the diameter, when you are choosing first point A, 3rd point is B. But when 1st point B then automatically it'll calculate 2nd point A. So AB is calculated twice.

What  nick1816  has done- He has chosen the diameter and worked forward so there was no repetition.
E.g. he chose AB first then the third point can be 6 ways. As we can have 4 diameters so 4*6 distinct rt. angled triangles.

firas92 · Answer

No of ways of choosing 3 points = 8C3=56

To get right angled triangles, two of our 3 points must be diametrically opposite so that they can form a right angle at the 3rd point

The first two points can be chosen in 4 ways.

Now we'll have 3 points on either side of the diameter so there are 6 ways of picking the 3rd point

Therefore, total number of favorable picks = 4*6 = 24

So, the probability that the three picked points form a right angled triangle = 24/56 = 3/7

abhishek31 · Answer

Hi , can anyone explain why the answer is 1/7 ?

Bunuel · Answer

_________________________
The OA is D. Edited. Thank you.

kitipriyanka · Answer

Eight points are equidistant on a circle. Hence there are 4 pairs of points which form a diameter.
In order to form a right-angled triangle, one of the sides should be diameter.
So, there are 4C1 ways to choose two points which form a diameter(as 4 such pair exists).

Now, once we have chosen 2 points, out of 6 remaining points on the circle, 3rd point can be any one of those 6 points. So there are 6C1 ways of choosing 3rd point

Now, total no of possible triangles is choosing any 3 points out of 8 points, which is 8C3.

Now probability = Possibility of forming a right angled triangle/ possibility of forming a triangle
=(4C1*6C1)/8C3
=3/7

Hence D

ScottTargetTestPrep · Answer

There are 8C3 = (8 x 7 x 6)/(3 x 2) = 8 x 7 = 56 ways to select 3 points from 8.

Now, let’s let the 8 points be A, B, C, D, E, F, G and H. Since these eight points are equally spaced on the circle. We see that each pair of consecutive points are separated by 360/8 = 45 degrees.

For any 3 chosen points to be the vertices of a right triangle, 2 of the vertices must be the endpoints of a diameter of the circle since the hypotenuse of a right triangle inscribed in a circle is a diameter of the circle. Therefore, if A is chosen, E has to be chosen, too. In other words, A and E need to be both chosen since AE is a diameter of the circle. If A and E are both chosen as the vertices of the triangle, then the last vertex can be any one of the remaining 6 points since the triangle will always be a right triangle once A and E are chosen. The same analogy can be made for B and F, C and G, and E and H since BF, CG and EH are diameters of the circle also.

Since there are 4 diameters (AE, BF, CG and EF) and each diameter has 6 ways to form a right triangle, there are 4 x 6 = 24 right triangles. Therefore, the probability that a triangle having the 3 points chosen as vertices will be a right triangle is 24/56 = 3/7.

Answer: D

Fdambro294 · Answer

The answer can be found by drawing the circle and labeling the 8 equidistant points A through H.

Since each arc is measure by the central angle, we’ll be able to connect 8 line segments to the center of the circle.

360/8 = 45 degrees = Each identical arc

Rule: the central angle is 2 times the inscribed angle subtended by the same arc - or, in other words - the inscribed angle formed at the circumference of the circle is (1/2) the central angle created by the same Arc

In order to have a 90 degree triangle with 3 points on the vertices (one of the angles as the Inscribed Angle), we need a central angle measure of 180 degrees ——-> diameter is the straight line, Central Angle equal to 180 degrees——-> since the diameter as a central angle is 180 degrees, the the inscribed angle subtended by the diameter is 90 degrees (1/2 * 180 deg.)

The above explanation leads to the rule that whenever an inscribed triangle has the diameter as one of its sides, it will be a right triangle.

Since each of the 8 points is equidistant from each other and the circle is symmetric, if we connect a pair of vertically opposite Points with any other remaining point ———-> one side of the inscribed triangle will be the diameter and we will have a Right Triangle

NUM = Number of Favorable Outcomes = no. of unique right triangles =

We will have 4 pairs of Veritically Opposite sides that will create a diameter across the circle (A-E) —- (B-F)——(C-G) ——(D-H)

We can pick any 1 of these 4 pairs in —-> “4 choose 1” = 4 ways

AND

for each pair chosen, there are 6 points remaining on the circle. If we connect any one of those pairs with one of the 6 remaining points, we will have a right triangle

(4) * (6) = 24 favorable outcomes.

DEN = Total possible outcomes = no. of ways to choose any 3 points out of a total of 8 points on the circle = “8 choose 3” = 8! / (3! 5!) = 56

Prob = 24/56 = 6/14 = 3/7

Answer: 3/7

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GMATE1 · Answer

Like most people I  solved this question using the combinatorial approach  (calculating how many option are desirable devided by how many options there)
=(4C1*6C1)/8C3 = 4*6/56 =3/7

BUT there is also a second approach     (the probability approach). In this approach you calculate how likely it is to make each choice. In this case this approach is a bit more difficult but generally I find this approach (at least in easy question) more intuitive.

We are picking 3 of 8 and we want to know what the probability is that we pick a two diagonals points and one other point. In other words the likelyhood of picking a pair and any other point and there are 4 pairs.

There are  three orders  inwhich ways we can pick a pair/the points of the diagonals in this case:

X1 & X2 representing the pair. Y representing any of the other 6 points)

1. X1 , X2, Y (first we pick the pair (X1 & X2) then any other of the 6 remaining points)
2. X1 , Y , X2 (first we pick one point of the pair (X1) then any other point (Y) and then we pick the second point of the pair (X2)
3. Y , X1, X2 (first we pick any of the 8 points (Y) and then we pick the pair (X1 & X2)

The  likelyhood of each of the three possible orders  is the following:

1. Choosing X1 first has a likelyhood of 1/8 we need to multiply this times the likelyhood of choosing X2 which is 1/7 and multiply this times the likelyhood of choosing any of the remaining numbers (likelyhood of Y) 6/6. The results is the likelyhood of choosing X1 first and X2 second (and Y third) but of course X1 and X2 could switch their order. Therefore we need to multiply the result times 2 and since there are four other pairs beside X1&X2 for which the likelyhood is the same we should multiply the result again times 4.

P1= 1/8 * 1/7 * 6/6 * 2 * 4 = 8/56 =  1/7

2. Choosing X1 first has a likelyhood of 1/8 we need to multiply this times the likelyhood of choosing Y second ,which is 6/7 and multiply this times the likelyhood of choosing X2 third, which has a likelyhood of 1/6.
The results is the likelyhood of choosing X1 first and X2 third but of course X1 and X2 could switch their order. Therefore we need to multiply the result times 2 and since there are four other pairs beside X1&X2 for which the likelyhood is the same we should multiply the result again times 4.

P2= 1/8 * 6/7 * 1/6 * 2 * 4 =  1/7

3.Choosing Y first has a likelyhood of 6/8 we need to multiply this times the likelyhood of choosing X1 second ,which is 1/7 and multiply this times the likelyhood of choosing X2 third, which has a likelyhood of 1/6.
The results is the likelyhood of choosing Y first and X1 second and X2 third but of course X1 and X2 could switch their order. Therefore we need to multiply the result times 2 and since there are four other pairs beside X1&X2 for which the likelyhood is the same we should multiply the result again times 4.

P3= 6/8 * 1/7 * 1/6 * 2 * 4 =  1/7

Adding the probabilites of the three probabilities we get the probability of Ptotal=1/7 + 1/7 + 1/7 = 3* 1/7 =  3/7

SaiVai · Answer

With the combinatorics method I got 24/56 i.e.3/7

With probability method, I got 1/7 as the answer.
as we have to choose three points of which two are diagonal and one is from the remaining 6
Ways of choosing the first point is 8/8
Ways of choosing Second point is 1/7
Ways of choosing Third point is 6/6

Therefore Probability of doing all these things together is 8/8 * 1/7 * 6/6 = 1/7. What am I missing here?

bumpbot · Answer

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Eight points are equally spaced on a circle. If 3 of the 8 points are

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