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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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 17 Jul 2008, 12:36
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 [ 16.13 KiB | Viewed 7816 times ]
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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Updated on: 17 Jul 2008, 14:47
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). 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.
Originally posted by tarek99 on 17 Jul 2008, 12:46.
Last edited by tarek99 on 17 Jul 2008, 14:47, edited 1 time in total.
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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17 Jul 2008, 13:18
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 2348 times ]
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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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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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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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 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
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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Updated on: 17 Jul 2008, 20:06
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
Originally posted by durgesh79 on 17 Jul 2008, 20:00.
Last edited by durgesh79 on 17 Jul 2008, 20:06, edited 1 time in total.
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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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
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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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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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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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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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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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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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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17 Aug 2018, 17:45
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.
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Re: An equilateral triangle ABC is inscribed in square ADEF, forming three
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