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28 Feb 2007, 00:16
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This one is from the OG. Please explain your solution in detail. Thanks!!
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28 Feb 2007, 03:00
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(D) for me 
At Q, we have a right angle. So, we can create all these rules:
o PQ^2 + QR^2 = (a+b)^2
o PQ^2 = a^2 + 2^2
o QR^2 = b^2 + 2^2
That implies,
a^2+4 + b^2+4 = (a+b)^2
<=> a^2 + b^2 + 8 = a^2 + 2*a*b + b^2
<=> a*b = 4
The diameter is equal to a+b
From 1
a=4, thus b=1 and a+b = 5
SUFF.
From 2
b=1, thus a=4 and a+b = 5
SUFF.
At Q, we have a right angle. So, we can create all these rules:
o PQ^2 + QR^2 = (a+b)^2
o PQ^2 = a^2 + 2^2
o QR^2 = b^2 + 2^2
That implies,
a^2+4 + b^2+4 = (a+b)^2
<=> a^2 + b^2 + 8 = a^2 + 2*a*b + b^2
<=> a*b = 4
The diameter is equal to a+b
From 1
a=4, thus b=1 and a+b = 5
SUFF.
From 2
b=1, thus a=4 and a+b = 5
SUFF.
28 Feb 2007, 19:58
Kudos
Bookmarks
There is another and less computive way to solve this.
Let X be point where perpendicular from Q meets PR
Then , we can easily prove, triangle PQX and QXR are SIMILAR triangles.
So if PX is twice QX then XR has to be half of QX,
alternatively if XR is half of QX then PQ has to be twice the IX
So both 1) and 2) are sufficient and thus answer is (D)
Let X be point where perpendicular from Q meets PR
Then , we can easily prove, triangle PQX and QXR are SIMILAR triangles.
So if PX is twice QX then XR has to be half of QX,
alternatively if XR is half of QX then PQ has to be twice the IX
So both 1) and 2) are sufficient and thus answer is (D)
