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# An equilateral triangle ABC is inscribed in square ADEF,

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17 Jul 2008, 12:46
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, 14:47, edited 1 time in total.
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17 Jul 2008, 13:18
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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 1401 times ]

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17 Jul 2008, 15:06
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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17 Jul 2008, 15:58
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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 1400 times ]

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17 Jul 2008, 16:02
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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17 Jul 2008, 20:00
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, 20:06, edited 1 time in total.
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14 Nov 2013, 13:29
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15 Mar 2015, 15:55
Hello from the GMAT Club BumpBot!

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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 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
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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
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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?

A. 4/3
B. $$sqrt(3)$$
C. 2
D. 5/2
E. $$sqrt(5)$$

Hi Bunuel, VeritasPrepKarishma, chetan2u,

Can you please provide some easy explanation for this question
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01 May 2016, 23:39
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?

A. 4/3
B. $$sqrt(3)$$
C. 2
D. 5/2
E. $$sqrt(5)$$

Hi Bunuel, VeritasPrepKarishma, chetan2u,

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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06 May 2016, 09:02
8 minutes, not worth it. Next!
Re: An equilateral triangle ABC is inscribed in square ADEF,   [#permalink] 06 May 2016, 09:02

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