Question on square? Help needed?

Thank You, Mathematishan, for the personal email with the invitation to answer. I am very glad to do so because this is a very well-known old problem, cited in many sources, and one of my favorites.

I know at least 3 solutions. One of them uses trigonometry, it is exactly what Mathsmanretired has submitted above. For the next two, follow the link below to see a picture:

http://farm5.static.flickr.com/4076/4742651609_439...

#1: Consider the given square ABCD along with another one, A'B'C'D', congruent to ABCD and corresponding sides parallel, and point O' in A'B'C'D', such that the triangle A'B'O' is equilateral.

The latter implies | A'O' | = | A'C' |, i.e. the triangle A'C'O' is isosceles with angles 30°, 75°, 75°. Hence the angle D'C'O' is 15° and C'D'O' is also isosceles. Now slide ABCD over A'B'C'D' (translation by a vector AA') until they coincide (A≡A', B≡B', C≡C', D≡D'). Since translation is isometry, equal angles OCD and O'C'D' will coincide, same about angles ODC and O'D'C', i.e. the above translation maps O into O'.

The latter implies triangles ABO and A'B'O' are congruent, but the second one is equilateral, so ABO is equilateral as required.

#2 (the most elegant): Draw a triangle BDP ≡ CDO as shown. The angle ODP is

90° - 2*15° = 60° and the above congruence implies | OD | = | PD |, hence the triangle ODP is equilateral. Consider triangle OPB.

We have | OP | = | PB | and the angle OPB is 360° - 150° - 60° = 150°,

hence triangles OPB and DPB are also congruent.

Now AOB is obviously isosceles with angle

ABO = 90° - 2*15° = 60°, i.e. equilateral.

This is actually not very easy. Draw a line through the middle of the square down through O. Let E be the mid-point of CD.

You should see straight away that OE/ED = tan15 ----> OE = (1/2)*tan15

Now tan15 can be found exactly in radical form by using tan30 = 1/sqrt3 and

tan2A = 2*tanA/(1 - (tanA)^2)

If you use A = 15 and t for tan15 in the above you get

1/sqrt3 = 2t/(1 - t^2) ----> t^2 + 2t*sqrt3 - 1 = 0

This can be solved for t by quadratic formula to get tan15 = 2 - sqrt3

Can you finish it from there? I think that it would be best if you do some it it yourself. Try to avoid the temptation to copy any complete solution which follows.

Set up a grid. You can put either on the top or side, I'll put RRYy on the top. Since all the R alleles are dominant (capital), all four columns will have a big R. Half of the Y alleles are dominant, half are recessive (because it's Yy). So you put two Y and two y in the square. Pretty much do the same thing for the RrYY but with R's and Y's switched since all Ys are dominant and only half the R's in this one. ... RY RY Ry Ry RY RY rY rY And fill out the blocks where they match up. There should be two of each letter in each box.

ans:

ABCD is square, so

AB=BC=CD=CA .........................1

in triangle OCD

OC=OD and angle OCD=ODC (as given)

that means triangle OCD is equilateral triangle ie all angle and arms are equal

so OC=CD=OD and angle OCD=ODC=COD

in triangle ABO and OCD

CD=AB

ie OC=OA and OA=OD

that means all arms of triangle ABO are same

ie proved that triangle ABO is an equilateral triangle

Here's some help:

A triangle's inner angles always add up to 180*

A square's inner angle is always 90*

its easy yar

draw the diagrm

pt o wl b at centre use theory of angles in a sqr

they r 90 fr adjact sides n centre fr diagonal

u wl solve it

al the bst