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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, 14:47, 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 1312 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]

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14 Nov 2013, 13:29

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Re: An equilateral triangle ABC is inscribed in square ADEF, [#permalink]

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15 Mar 2015, 15:55

Hello from the GMAT Club BumpBot!

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An equilateral triangle ABC is inscribed in square ADEF, [#permalink]

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16 Mar 2015, 06:04

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

Re: An equilateral triangle ABC is inscribed in square ADEF, [#permalink]

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25 Apr 2016, 08:14

just skip the question at the real exam! it takes a while just to figure out how to place the triangle into the square and then the whole work is about algebraic manipulation involving quadtatics. and most important - picking smart numbers just won't work here. a great question to waste >3 min and mental energy and get it wrong _________________

An equilateral triangle ABC is inscribed in square ADEF, [#permalink]

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01 May 2016, 07:37

tarek99 wrote:

An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?

An equilateral triangle ABC is inscribed in square ADEF, [#permalink]

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01 May 2016, 23:39

Expert's post

smartguy595 wrote:

tarek99 wrote:

An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?

Can you please provide some easy explanation for this question

Hi,

Here is a solution for the Q..

See the attached figure-

let the sides be 4 and BE be x, than BD = 4-x.. Now we have made a equilateral triangle .. so its side from triangle BCE = \(\sqrt{x^2+x^2}\) = \(\sqrt{2}x\)..

Now look at the right angle triangle ABD.. here AB^2 = AD^2 + BD^2.. \((\sqrt{2}x)^2 = 4^2 + (4-x)^2\)... \(2x^2 = 16 + 16 + x^2 - 8x..\) \(x^2 + 8x - 32 = 0\).. Roots of Quad eq ax^2+bx+c=0 are \(-b+- \sqrt{b^2-4ac}/2\).. the VALID value of x comes out as \(4(\sqrt{3}-1)\)..

Now area of ABD = \(\frac{1}{2} * 4 * (4-x) = 2*(4 - 4(\sqrt{3}-1))\) .. => \(2*(8-4\sqrt{3}) = 8(2-\sqrt{3})\)..

Area of BCE = \(\frac{1}{2} *x*x = \frac{1}{2} (4(\sqrt{3}-1))^2 = \frac{1}{2}*16*(\sqrt{3}-1)^2 = 8(3+1-2\sqrt{3})\).. =>\(8(4-2\sqrt{3})= 16(2-\sqrt{3})\).. so the ratio of areas of \(BCE/ABD = 16(2-\sqrt{3})/8(2-\sqrt{3})= 2\) C

Ofcourse there is a geometrical approach too.. Try it

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