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# Tough Geometry problem

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16 Nov 2010, 21:03
00:00

Difficulty:

(N/A)

Question Stats:

67% (00:07) correct 33% (00:11) wrong based on 10 sessions

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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.

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17 Nov 2010, 17:09
1
KUDOS
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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17 Nov 2010, 11:00
Since you posted this on PS forum, could you give the answer choices as well?

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18 Nov 2010, 03:26
The answer given was 1:3, I could not reckon how though !
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Kudos [?]: 607 [0], given: 40

Intern
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GPA: 3.29
WE: Engineering (Consulting)

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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 6044 times ]

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24 Aug 2015, 17:11
I thought this problem was really hard, but actually it's really really simple.

We know that triangle ABC is a 3-4-5 triangle. simple from knowing AC = 3 and CB = 4

CB = 4 = CD + DB

Equation 1 : CD + DB = 4

We are told that P(ABD) = P(ADC)

Equation 2 : CD - DB = 2

CD + DB = 4
CD - DB = 2
2CD = 6

CD = 3

Therefore DB = 1

And finally CD:DB > 3:1 or whatever direction the question asked.

Essentially you don't need to know AD.

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13 Sep 2016, 03:33
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08 Nov 2017, 12:11
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Tough Geometry problem   [#permalink] 08 Nov 2017, 12:11
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