The diagram above shows 3 identical circles placed inside a right tria

Question

https://gmatclub.com/forum/download/file.php?id=53274 The above diagram shows 3 identical circles placed inside a right triangle so that all 3 circles are tangent to line AB. What is the radius of each circle? A) 5 B) 6 C) 7 D) 8 E) 10 rnMKbFE.png

BrentGMATPrepNow · Accepted Answer

Key property: A line tangent to a circle is perpendicular to the circle's radius  at the point of intersection

First, if we add the centers of each triangle as as well as the point E, we get the following: 
 https://i.imgur.com/RRFHjyp.png

At this point, if we draw lines from point E to each of the three vertices, look at the following:
 https://i.imgur.com/M9lyRCF.png

At this point we may recognize that: ( area of ∆AEB ) + ( area of ∆CEA ) + ( area of ∆CEB ) =  area of ∆ACB

Area of triangle = (base)(height)/2

So, we get:  (40)(r)/2  +  (30)(5r)/2  +  (50)(r)/2  =  (40)(30)/2  
Simplify to get: 20r + 75r + 25r = 600
Simplify: 120r = 600
Solve: r = 600/120 = 5

Answer: A

Cheers,
Brent

sujoykrdatta · Answer

Referring to the diagram below: 111.JPG RQ is drawn tangent to the 3rd circle as shown Assuming each circle to be of radius r, we have: AR = 4r = PQ => RB = 40 - 4r Triangles CPQ and CAB are similar => CP/CA = PQ/AB = CQ/CB => CP/30 = 4r/40 = CQ/50 => CP = 3r and CQ = 5r => BQ = 50 - 5r; RB = 40 - 4r; AP = RQ = 30 - 3r From O (center of the 3rd circle), the radii OT and OU are joined as shown RT = OU = OT = RU = r => UB = RB - RU = 40 - 4r - r = 40 - 5r => BS = BU = 40 - 5r (tangents from the same point to a circle are equal) => QS = BQ - BS = (50 - 5r) - (40 - 5r) = 10 => QT = QS = 10 (tangents from the same point to a circle are equal) Thus: QR = QT + TR => 30 - 3r = 10 + r => r = 5 A couple of questions: 1. Why ask for the GREATEST possible radius, when the exact radius can be solved for? 2. Does the question ask for the diameter?

BrentGMATPrepNow · Answer

Good questions
1) I should have just asked for the radius. That said, the answer is 80/11 either way.
I've edited the question to just ask for the radius.

2) In my original "solution" I said the length of blue dotted line was 2.5r when it's clearly 5 radii (i.e.. 5r)

I've edited both my solution AND the answer choices. 
Good catch!!

Cheers and thanks,
Brent

GMATWhizTeam · Answer

Solution

• Let the radius of each circle be r.
• It is given that AB is tangent to all three identical circles, So, radius of each circle will be perpendicular to AB at the point of tangency. 
 o So, DE is perpendicular to AB.
o Also,  AG = DE = r 
  Thus, as shown in the below diagram, AEDG is a rectangle, 
 https://www.dropbox.com/s/sovl48zoc5zy65p/trianlge_12_03.png?dl=1 
 • And,  AE = GD = 5r 
• Also,  ∠CGD = 90 °, and triangle CGD is a right-angled triangle.   
• Now, using Pythagoras theorem in triangle, CGD, we get:
 o  CD^2 = CG^2 + GD^2 ⟹ CD^2 = (30-r)^2 + (5r)^2 …………Eq.(i)  
• Now, in the right-angled triangles DEB and DFB,
 o Side  DE = DF = r  and 
 o hypotenuse is common, i.e. DB
o So, EB must be equal to FB i.e.  EB = FB = 40 – 5r   
• So,  FC = BC -BF = 50 – (40 – 5r) = 10+ 5r  
• Now, using Pythagoras theorem in triangle in triangle, CDF,
 o  CD^2 =DF^2 +FC^2 ⟹ CD^2 = (10+5r)^2 + (r)^2 ……………….Eq.(ii)  
• From Eq.(i) and (ii), we can write,
 o  (30-r)^2 + (5r)^2 = (10+5r)^2 + (r)^2 
o Or,  900 + r^2 – 60r + 25r^2 = 100 + 25r^2 + 100r + r^2 
o Or,  160r = 800 ⟹ r = 5   
Thus, the correct answer is  Option A.

bumpbot · Answer

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The diagram above shows 3 identical circles placed inside a right tria

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