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

Question

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

alfyG · Accepted Answer

My solution is very similar to yours - I got 2 as well. Maybe a bit shorter. I am attaching my solution as a jpeg file. Tell me what you think.

vshaunak@gmail.com · Answer

It should be 'C'.
See the attached solution.

maratikus · Answer

let's denote AB = BC = AC = 1. B belongs to DE, C belongs to EF
angle BAD = 15 degrees, EBC = 45 degrees
let's draw DH: H belongs to AB and DH is perpendicular to AB -> S(ABD) = 1/2*(AB)*(DH) = 1/2*1*(sin 15*cos15) = 1/2*(1/2)*sin(30) = 1/8 //DH = AB*sin(BAD)*cos(BDH) = AB*sin(BAD)*cos(BAD) = 1/2*sin(2*BAD)

let's draw EN: N belongs to BC and EN is perpendicular to BC -> S(BEC) = 1/2*(BC)*(EN) = 1/2*1*(sin45*cos45) = 1/2*1/sqrt(2)*1/sqrt(2)= 1/4

S(BEC)/S(ADB) = 2 -> C

abhitheCEO · Answer

Let's denote AB = BC = AC = 1. B belongs to DE, C belongs to EF
angle BAD = 15 degrees, EBC = 45 degrees
Assume the length of the side of the square as y and BE = x.
Now, A(BEC)/A(ADB) =(AD* DB)/ (BE*CE) ignoring 1/2 in denominator as well as numerator
as BE = CE = x and AD = y and DB = (y-x)
if you replace values and on solving, A(BEC)/A(ADB) = x^2 / (y^2 -xy)

Now, as per pythagorus, BC = AB = 2 * sqrt x
and AB^2 = AD^2 + DB^2
put values for AB = 2 * sqrt x and AD = y and DB = y-x
solve, you'll get x^2 / (y^2 -xy) = 2

Hence, c

goalsnr · Answer

Let Y be length of the square
ABC is a triangle such that B intersects DE and C intersects FE (This info is from problem description)

Since it is an quilateral triangle , each interior angle of triangle ABC is 60 degrees.
Therefore angle BAD = angle CAF = 15. 
Tiangle AFC is right angled. So angle ACF = 75.

Now solve angles for triangles ADB abd BCE.

Tiangle BCE turns out to be an isocles right angled triangle.

Area of BCE = BE * CE/2

let BE =x

Area of BCE = x* x/2

Area of ADB = AD* DB/2

but DB+BE=y

->DB = y-x

Area of ADB = y* (y-x)/2

Since ADB is a right angled triangle.

y^2 +(y-x)^2 = 2 * x^2

->y^2 -xy = x^2/2

Area of BCE/Area of ADB =  / 
 = x* x/y* (y-x) =2

walker · Answer

Fig

walker · Answer

By the way, second approach:

let x - a side of the equilateral triangle ABC
y - a side of the squre

1. The diagonal of squre =  \sqrt{2}y = \frac{\sqrt{3}}{2}x+\frac{1}{2}x=\frac{x}{2}*(\sqrt{3}+1) 
2. The area of squre =  y^2=\frac{x^2}{8}*(\sqrt{3}+1)^2=\frac{x^2}{8}*(4+2\sqrt{3})=\frac{x^2}{4}*(2+sqrt{3}) 
3. The area of triangle ABC =  \frac12*\frac{\sqrt{3}}{2}x*x=\frac{\sqrt{3}}{4}x^2 
4. The area of triangle BEC =  \frac12*\frac{x}{2}*x=\frac{x^2}{4} 
5. The area of triangle ADB = 1/2 * (The area of squre - The area of triangle ABC - The area of triangle BEC) =  \frac12*(\frac{x^2}{4}*(2+sqrt{3})-\frac{\sqrt{3}}{4}x^2 -\frac{x^2}{4})=\frac{x^2}{8} 
6. ratio =  \frac{\frac{x^2}{4}}{\frac{x^2}{8}}=2

marcodonzelli · Answer

Since |AB|=|BC|, x2 + 2xy + y2 +x2 = 2y2, so 2x2 +2xy =2x(x+y)= y2
The ratio of the area of triangle BEC to that of triangle ADB is y2/x(x+y) =2

AB and BC obviously are 2 sides of the triangle. x and y are the 2 parts of the side of the square

KarishmaB · Answer

You need to get y in terms of s to get the ratio. 
y - side of square
s - side of equilateral triangle

Note that the diagonal of a square is  \sqrt{2}*side = \sqrt{2}*y

Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC

Altitude of equilateral triangle is given by  \sqrt{3}/2 * Side = \sqrt{3}/2 * s 
Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2

So  \sqrt{2}*y = \sqrt{3}/2 * s + s/2 
So  y = s * (\sqrt{3} + 1)/(2\sqrt{2})

Now, area of square =  y^2 = s^2(\sqrt{3} + 1)^2/8

Area of triangle ABC =  (\sqrt{3}/4) * s^2

Area of triangle BEC  = s^2/4

Area of triangle ADB = Area of triangle AFC  = 1/2 * (s^2(\sqrt{3} + 1)^2/8 - (\sqrt{3}/4) * s^2 - s^2/4) = s^2/8

Hence the required ratio is  (s^2/4)/(s^2/8) = 2

jlgdr · Answer

Karishma this is an awesome explanation for this tough problem, thanks a lot Just one small doubt, how do you know that ADB and AFC right triangles are equal? and both different from BCE, I have a bit of trouble making those sort of observations Many thanks! Cheers! J

KarishmaB · Answer

Regular polygons inscribed in other regular polygons or circles make symmetrical figures. Try drawing them out. The sides/angles you think are equal will usually be equal.
Angle DAF is 90 and BAC is 60 so angles DAB and FAC will be 15 degrees each. Also AD = AF and AB = AC. So triangles ADB and AFC are congruent. On the other hand, in triangle BEC, side BE = CE so it is a 45-45-90 triangle.

GoonerGalactico · Answer

Either the question is wrong or this diagram. I don't see ACF as a right angle.

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

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