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# An equilateral triangle ABC is inscribed in square ADEF,

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Re: PS: Geometry [#permalink]  17 Jul 2008, 11:46
jallenmorris wrote:
Here is a solution that works pretty quickly, but I'm not sure about finishing it to come up with an actual ratio of 2:1.

See Picture Attached (same as other one, but this post is on page 2, so i attached it again).

Right now we have BEC = \frac{y^2}{2} and ADB = 1*(1-y)*0.5 or \frac{1-y}{2}

This is the same as \frac{y^2}{2} divided by \frac{1-y}{2} or \frac{y^2}{2} * \frac{2}{1-y}=\frac{y^2}{1-y}

We know that y is a fraction because we made the entire side = 1, but how much of that side is y?

Attachment:
TriangleInscribed.jpg

I guess we have to show a relationship between AD, DB, and AB such as (AD)^2 + (DB)^2 = (AB)^2 just as how we've established that relationship between BE, EC, and BC such as (BE)^2 + (EC)^2 = (BC)^2 but that will consume time.

Last edited by tarek99 on 17 Jul 2008, 13:47, edited 1 time in total.
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Re: PS: Geometry [#permalink]  17 Jul 2008, 12:18
I think I have it figured out:

If area of Tri. BEC = \frac{y^2}{2} and Tri.ADB = \frac{1(1-y)}{2} then we have
\frac{y^2}{2} * \frac{2}{1-y} = \frac{y^2}{1-y} but we have to figure out what this is in numbers.

So

Notice that the hypotnuse of ADB and BEC are both Z

so...
y^2 + y^2 = z^2 and 1^2 + (1-y)^2 = z^2..so
y^2 + y^2 = 1^2 + (1-y)^2
2y^2 = 1 + 1 - 2y + y^2
2y^2 = 2 - 2y + y^2
y^2 = 2 - 2y
y^2 = 2(1 - y) - now multiply both sides by 1/2
\frac{y^2}{2} = 1 - y - now multiply both sides by \frac{1}{y^2}
\frac{1}{2} = \frac{1 - y}{y^2}

Notice above, that the ratio is \frac{y^2}{1-y}, so if \frac{1-y}{y^2}=\frac{1}{2}, then the inverse \frac{y^2}{1-y}= 2 or answer C!!

Wow, and it only took us all day!

Attachment:

TriangleInscribed.jpg [ 16.13 KiB | Viewed 588 times ]

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Re: PS: Geometry [#permalink]  17 Jul 2008, 14:06
jallenmorris wrote:
I think I have it figured out:

If area of Tri. BEC = \frac{y^2}{2} and Tri.ADB = \frac{1(1-y)}{2} then we have
\frac{y^2}{2} * \frac{2}{1-y} = \frac{y^2}{1-y} but we have to figure out what this is in numbers.

So

Notice that the hypotnuse of ADB and BEC are both Z

so...
y^2 + y^2 = z^2 and 1^2 + (1-y)^2 = z^2..so
y^2 + y^2 = 1^2 + (1-y)^2
2y^2 = 1 + 1 - 2y + y^2
2y^2 = 2 - 2y + y^2
y^2 = 2 - 2y
y^2 = 2(1 - y) - now multiply both sides by 1/2
\frac{y^2}{2} = 1 - y - now multiply both sides by \frac{1}{y^2}
\frac{1}{2} = \frac{1 - y}{y^2}

Notice above, that the ratio is \frac{y^2}{1-y}, so if \frac{1-y}{y^2}=\frac{1}{2}, then the inverse \frac{y^2}{1-y}= 2 or answer C!!

Wow, and it only took us all day!

Attachment:
TriangleInscribed.jpg

yeah, I just got the answer the same way you did. wow, this is such a ridiculous problem. It also took me the whole day just dedicating to this one annoying question. can you imagine seeing a question like that on the real gmat??? This question was provided by the GMATprep, so you better take this question seriously! heheh....great job! now I can go to bed so that I can dream about this horrifying experience....heheh...
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Re: PS: Geometry [#permalink]  17 Jul 2008, 14:58
4
KUDOS
Since |AB|=|BC|,x^2 + 2xy + y^2 +x^2 = 2y^2, so 2x^2 +2xy =2x(x+y)= y^2
The ratio of the area of triangle BEC to that of triangle ADB is y^2/x(x+y) =2
Attachments

Dibujo12.jpg [ 10.28 KiB | Viewed 592 times ]

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Re: PS: Geometry [#permalink]  17 Jul 2008, 15:02
jallenmorris wrote:
Attachment:
TriangleInscribed.jpg

I really think that the poster should post the actual diagram..this problem isnt as hard if the diagram is given...

Ok lets see..

side of Square=S...side of Equilatral=A..

suppose point B and C are X..(btw you will quickly see that B and C are exactly X away)..i.e BE=CE..

so A^2=S^2+(s-x)^2

similary BC=A => A^2=x^2+x^2 or 2x^2

area of ADB=1/2 (S)*(s-x) area of BCE=1/2 (X)*(X)

Ok, so we S^2 - 2xS + x^2=2X^2
x^2=2s^2-2xs or 2(s^2-xs)

area of BCE=1/2*x^2 we know x^2 interms of S..

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Re: PS: Geometry [#permalink]  17 Jul 2008, 19:00
jallenmorris wrote:
durgesh, can you label which of these goes to which triangle and how you came up with \frac{\sqrt{3}}{4}?

\frac{\sqrt{3}}{4} * 2a^2

1/2 a^2

2 * 1/2 * b(a+b)

\frac{\sqrt{3}}{4} * 2a^2 - Equilateral triangle ABC with side a * sqrt2

1/2 a^2 - trinagle BCE

2 * 1/2 * b(a+b) - traingle ADB + traingle ACF

Last edited by durgesh79 on 17 Jul 2008, 19:06, edited 1 time in total.
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Re: An equilateral triangle ABC is inscribed in square ADEF, [#permalink]  14 Nov 2013, 12:29
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Re: An equilateral triangle ABC is inscribed in square ADEF,   [#permalink] 14 Nov 2013, 12:29
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