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# Tough Geometry problem

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SVP
Joined: 16 Nov 2010
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Tough Geometry problem [#permalink]  16 Nov 2010, 21:03
00:00

Difficulty:

(N/A)

Question Stats:

60% (01:04) correct 40% (00:11) wrong based on 3 sessions

A right angled triangle ABC was given in which angle C is 90 degrees. BC=4, AD=3. D is a point on BC. Given that the perimeter of triangle ABD is equal to the perimeter of triangle ADC. Then find the ratio BD/DC.
Intern
Joined: 17 Aug 2009
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Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V42
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Kudos [?]: 11 [0], given: 4

Re: Tough Geometry problem [#permalink]  17 Nov 2010, 11:00
Since you posted this on PS forum, could you give the answer choices as well?
Intern
Joined: 17 Aug 2009
Posts: 41
Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V42
GPA: 3.29
WE: Engineering (Consulting)
Followers: 0

Kudos [?]: 11 [0], given: 4

Re: Tough Geometry problem [#permalink]  17 Nov 2010, 17:09
chaoswithin wrote:
Since you posted this on PS forum, could you give the answer choices as well?

The reason why I ask is because I can get a range for the possible answer, but not an exact number.

To get the upper limit, we can use the fact that the perimeters of the two triangles are equal and since they share the side AD, we can see that AB+BD = AC + DC

and since AB has to be > AC, BD has to be < DC so BD/DC has to be < 1.

for the lower limit, since AD = 3, DC has to be < 3. and since BD + DC = 4, when DC = 3 BD = 1 and BD/DC has to be > 1/3.

Therefore, 1/3 < BD/DC < 1. This is the best I can do with the givens.

If anyone else could shine some light on this problem, I would appreciate it.
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Re: Tough Geometry problem [#permalink]  18 Nov 2010, 03:26
The answer given was 1:3, I could not reckon how though !
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Intern
Joined: 17 Aug 2009
Posts: 41
Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V42
GPA: 3.29
WE: Engineering (Consulting)
Followers: 0

Kudos [?]: 11 [0], given: 4

Re: Tough Geometry problem [#permalink]  18 Nov 2010, 08:39
subhashghosh wrote:
The answer given was 1:3, I could not reckon how though !

Using the givens, the triangle should look similar to the triangle in the attachment.

In order for BD/DC to be 1/3, BD = 1 and DC = 3.

But since triangle ADC has hypotenuse AD = 3. DC cannot equal 3.

In order for BD/DC to be 1/3, the original question should be rephrased so AC = 3 instead of AD = 3.

If AC = 3.

AB = 5 (b/c the triangle becomes 345 triangle).

and we are left with two equations and two unknowns where

AB + BD = AC + DC and BD + DC = 4.

which results in BD = 1 and DC = 3.
Attachments

aa.JPG [ 5.66 KiB | Viewed 3044 times ]

Re: Tough Geometry problem   [#permalink] 18 Nov 2010, 08:39
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