Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 May 2017, 07:19

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m21 #20

Author Message
Intern
Joined: 13 Oct 2010
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 3

### Show Tags

09 May 2011, 04:39
The distance between points is sqrt10, sqrt10, sqrt20. This is 45, 45, 90 triangle. By circle properties if any inscribed triangle has 90 degree angle its hypotenuse passes through its center. So diameter is hypotenuse of triangle =sqrt20.
Intern
Joined: 27 Apr 2012
Posts: 12
Schools: Sloan '17 (WL)
GMAT 1: 670 Q47 V35
GPA: 4
WE: Supply Chain Management (Consumer Products)
Followers: 0

Kudos [?]: 1 [0], given: 131

### Show Tags

14 May 2012, 07:09
newdawn wrote:
My approach would be:
Let x,y is the center of the circle, then distance between center and any point on the circle is same equal to R.
Given that (1,2), (2,5) and (5,4) lie on circle, we can have following equations:

$$(x-1)^2 + (y-2)^2 = (x-5)^2 +(y-4)^2$$
solve it for
$$2x+y=9$$

Next take (2,5) and (5,4)
$$(x-2)^2 + (y-5)^2 = (x-5)^2 +(y-4)^2$$
and you will get
$$3x-y=6$$

Solve for x and y center is (3,3) and distance i.e. Radius is $$\sqrt{(5-3)^2+(4-3)^2} = \sqrt{5}$$

So Diameter is $$\sqrt{20}$$

I did the same. However it's way too long and complex calculation, gmat suggests easier solutions to be within 2 minutes limit
Intern
Joined: 07 Mar 2013
Posts: 6
Concentration: Entrepreneurship, Social Entrepreneurship
Followers: 0

Kudos [?]: 0 [0], given: 12

### Show Tags

13 May 2013, 09:09
In case the triangle ABC is not a right angled triangle, we can find the diameter using the following general formula:

Diameter = abc/(2*Area(ABC))

a,b,c are the sides opposite to angles A, B and C respectively.

So if we have the coordinates of all the vertices of the triangle we can easily find the diameter using the above formula
Current Student
Joined: 11 Apr 2013
Posts: 53
Schools: Booth '17 (M)
Followers: 0

Kudos [?]: 27 [0], given: 13

### Show Tags

13 May 2013, 10:11
Just want to note that I've seen a lot of people throwing around the "rule" that if a triangle is inscribed in a circle, then it either a) must be a right triangle or b) have it's hypotenuse pass through the circle's center. Neither of these are true...

The rule is that IF a right triangle is inscribed in a circle, then it's hypotenuse passes through the center of the circle (and is thus a diameter). You can inscribe a virtually infinite number of triangles in a circle which would not be right triangles and would thus not pass through the circle's center.

So to do this problem you must first determine that the triangle formed by the points is indeed a right triangle. Using (Y1-Y2)/(X1-X2) to find the slopes of the three line segments, you will see that segments intersecting at (2,5) have inverse and opposite slopes (3, -1/3) and are thus perpendicular to each other and meet at a right angle. From there it is as simple as applying the pythagorean theorm to find the long side, in this case x^2 = 4^2 + 2^2.

It seems like some people may have gotten "lucky" assuming that the long side was the hypotenuse of a right triangle, and thus a diameter (and the answer) - Just wanted to note that wouldn't always be the case
Intern
Joined: 12 Sep 2012
Posts: 16
Followers: 0

Kudos [?]: 3 [0], given: 10

### Show Tags

13 May 2013, 22:43
How we did know that the line is diameter?
Intern
Joined: 02 Oct 2012
Posts: 13
Location: United States
Concentration: Entrepreneurship, General Management
GMAT 1: 720 Q49 V38
GPA: 2.77
WE: General Management (Education)
Followers: 0

Kudos [?]: 6 [0], given: 12

### Show Tags

15 May 2013, 02:34
Good Question. For a sec I thought it is a tough question till I worked out the length of the sides.

The length of the sides work out to \sqrt{10} \sqrt{10} and \sqrt{20}

It implies that it is an isosceles triangle with a right angle.
All triangles with base as diameter and the vertex touching a point on circumference of the circle - the vertex angle is 90 degree.

So it works out. The base with side \sqrt{20} is the diameter of the circle as well.
Current Student
Joined: 28 Feb 2013
Posts: 15
Location: United States
Concentration: Statistics, Strategy
GMAT Date: 03-12-2014
GRE 1: 324 Q154 V170
WE: Information Technology (Other)
Followers: 1

Kudos [?]: 12 [0], given: 18

### Show Tags

29 Apr 2014, 16:04
deepak4mba wrote:
Guys, correct me if I am wrong in my logic.
Ignore the point (2,5). Considering the fact that the circle touches all the points, the line joining the (1,2)--say point A and (5,4)---say point B is the diameter(AB).
Now, draw perpendiculars from A and B across the coordinates which will join at Point C---(5,2).
Triangle, thus, formed is right angled triangle- triangle ABC.
Using simple coordinate numbers, the length of AC = 4 and length of BC = 2.
Applying Pythogoras Theorem, the diameter i.e. the hypotenuse of the triangle ABC = sqrt (20).

draw the above figure with simple coordinates to follow the above logic.

this is exactly how i did it.

Its much easier to visual if you have either a horizontal and/or vertical line in your triangle.

I drew dotted lines from points (5,4) and (1,2) to point (5,2) creating a right triangle angle at that new point with a shared hypotenuse with the original triangle. The length of the shared hypotenuse is what we are looking for.

side x which goes from point (1,2) to (5,2) has length 4
side y which goes from point (5,4) to (5,2) has length 2

a^2 + B^2 = c^2
c is the value we are looking for as length c is equal to side z the shared hypotenuse.

4^2 + 2^2 = c^2 16+4 = sqrt (20)

Q.E.D.
Re: m21 #20   [#permalink] 29 Apr 2014, 16:04

Go to page   Previous    1   2   [ 27 posts ]

Similar topics Replies Last post
Similar
Topics:
M21 - #2 1 30 Sep 2009, 07:50
23 m21 #8 21 17 Sep 2012, 08:06
18 M21#29 22 30 Jun 2013, 14:21
13 M21#15 16 15 Mar 2013, 20:30
32 M21 #28 26 13 Jun 2014, 07:23
Display posts from previous: Sort by

# m21 #20

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.