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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).
We're looking for BEC:ADB.
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.
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 1005 times ]
_________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
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...
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 \(y^2/x(x+y) =2\)
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]
15 Mar 2015, 14:55
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An equilateral triangle ABC is inscribed in square ADEF, [#permalink]
16 Mar 2015, 05:04
jallenmorris wrote:
Here is how I approached the problem.
AD (side of square) = 1 BE (side of 45:45:90 triangle) = y CB (side of equilateral triangle) = z
We're looking for the ratio of the area of BEC to ADB, or \(\frac{BEC}{ADB}\)
If area of Tri. BEC = \(\frac{y^2}{2}\) and Tri.ADB = \(\frac{1(1-y)}{2}\) then we have \(\frac{y^2}{2} * \frac{2}{1-y} = \frac{y^2}{1-y}\)
if y = \(\frac{1}{2}\), then y^2 = 1/4 and 1-y = 1/2, so we get 1/4 : 1/2 or 1/2....that's not an option.
Attachment:
TriangleInscribed.jpg
We can't take any value for y. There's only one way to put an equilateral triangle inside a square if we fix the square side, so if we chose that side then y is not flexible, and if you try for y=1/3 you'll not find the correct ratio for example.
I have a solution: Let's take a=1 as the side of the square, and let's name BE=x. Area of ADB = 1.(1-x)/2=(1-x)/2 Area of BEC=x^2/2 Therefore, ratio = x^2/(1-x)
Let's find x:
AE=sqrt(2)
Side of the Equilateral Triangle = x*sqrt(2) because BEC is isosceles.
On the other hand, if we apply Pythagoras on triangle ADB: (1-x)^2 = (x*sqrt(2))^2 -1 = 2x^2-1 which leads to a quadratic equation with one positive solution: x=sqrt(3) - 1
Let's then replace x with its value in ratio = x^2/(1-x) --> ratio= 2
gmatclubot
An equilateral triangle ABC is inscribed in square ADEF,
[#permalink]
16 Mar 2015, 05:04
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