An equilateral triangle ABC is inscribed in square ADEF, forming three

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.

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!

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...

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 ADB =1/2* (s^2-xs)

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

ADB/BCE = (s^2-xs)/2(s^2-xs)=1

ratio of BCE to ADB=2

took about 2 mins

\(\frac{\sqrt{3}}{4}\) * 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

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

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

Hi Bunuel, VeritasPrepKarishma, chetan2u,

Can you please provide some easy explanation for this question
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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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Hi Bunuel!
Could you plz explain how to establish that triangles ACF and ADB are congruent?
We have two sides(AF=AD & AC=AB) and one non-included angle (angles AFC & ADB) same in both the triangles.
I understand that it is none of the cases from SSS,SAS,ASA or AAS.

Very tricky question. The actual setup of the diagram took a bit of time to figure out.

Area of BEC = y^2/2
Area of ADB = 1(1-y)/2

Need = Area of BEC/Area of ADB = y^2/(1-y)

AB = BC (Sides of equilateral triangle)

1 + 1 + y^2 - 2y = 2y^2
2 + 2y = y^2
2 (1 - y) = y^2
y^2/(1-y) = 2

ANSWER: C

Why is DB = CF ?

Its a congruent triangle

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?


Good Question +1.

Let's discuss the step-by-step approach to solve this Question.

Step 1: Draw the figure w.r.t to that data given in the question. The naming of the figure should be as per the info given. Please refer to Fig. 1 in the attachment.



Step 2: Use the information given and try to figure out the unknown sides.

ADEF is a square. So, let's say the side AD = AF = FE = ED = a .

Since △ ABC is an equilateral triangle, AB = BC = AC

If you analyze the △ABD and △AFC i.e △1 and △2 in the fig.2, they are 2 right-angled triangles with the same hypotenuse ( AB = AC) and one of the sides is also equal (AD = AF = a).

If the hypotenuse and a side of a right-angled triangle are equivalent to the hypotenuse and a side of the second right-angled triangle, then the two right triangles are said to be congruent. So, we can conclude that the other side of the triangles is also equal. i.e DB = FC .

Let's say DB = FC = b. Then, BE = a - b , also CE =a-b. Please refer the figure 2 for better understanding.

Step 3: We are asked to find the ratio of the area of △ BEC to that of △ADB?

Area of a triangle = 1/2 * base * height

Area of △ BEC = 1/2 * (a-b) * (a-b)

Area of △ ADB= 1/2 * a * b

Area of △ BEC/Area of △ ADB = (1/2 * (a-b) * (a-b))/(1/2 * a * b) = \((a-b)^2/ab\)

Since there are 2 variables in this ratio, we need to convert them to a single variable in order to get the value for this ratio.

For that, we need to find a relationship between a and b.
Analyzing △ 1 and △ 3 in figure 3, we can see that their hypotenuse are equal (△ ABC is equilateral, AB =BC )

Can we equate the hypotenuse in both triangles to find the connection between a and b? Absolutely yes.

\(AB^2 = BC^2\)

In △ 1, \(AB^2\) = \(a^2 + b^2\) and in △ 2, \(BC^2 = (a-b)^2 + (a-b)^2\)

Equating both, \( a^2 + b^2\) = \((a-b)^2 + (a-b)^2\)

\(a^2 + b^2 = 2(a-b)^2\)

In the above equation, we can replace \(a^2 + b^2\) by \((a-b)^2 + 2ab\) ( Using algebraic identities)

\((a-b)^2 + 2ab = 2(a-b)^2\)

\((a-b)^2 = 2ab\)

Area of △ BEC/Area of △ ADB = \((a-b)^2\)/ab. = 2ab/ab = 2

Option C is the answer.

Thanks,
Clifin J Francis,
GMAT Quant Mentor
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