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Geometry Level 4

Given a parallelogram whose acute angle is 6 0 60^{\circ} . Find the ratio of the lengths of the sides of the parallelogram if the squares of the lengths of the diagonals have a ratio of 1 : 3 1:3 .


The answer is 1.

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3 solutions

Ahmad Saad
Jun 5, 2016

Very good use of Cosine rule \text{Very good use of Cosine rule} , + 1 +1 !

Rishabh Tiwari - 5 years ago

Solution using complex numbers:

Let one side of the parallelogram be represented by a a ( a C a \in \mathbb{C} ).

Then the other side is a r e i π 3 are^{i\frac{\pi}{3}} ( r R + r \in \mathbb{R}^+ ) where r r is the ratio of the sides.

The given data is equivalent to the following equation: a ( 1 + r e i π 3 ) ) 2 a ( 1 r e i π 3 ) ) 2 = 3 \frac{|a(1+re^{i\frac{\pi}{3}}))|^2}{|a(1-re^{i\frac{\pi}{3}}))|^2} = 3

Solving this equation, we get r = 1 \boxed{r=1}

AWESOME SOLUTION ! \text{AWESOME SOLUTION !} +1 !

Rishabh Tiwari - 5 years ago
Mehul Arora
Jun 4, 2016

A parallelogram whose acute angle is 6 0 60^{\circ} is a Rhombus.

Why is that necessary?

Rishabh Tiwari - 5 years ago

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I tried to minimize my solution. Adding a proof is easy. I can certainly add a proof.

Mehul Arora - 5 years ago

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Ok then please add a proof . Thank you.

Rishabh Tiwari - 5 years ago

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@Rishabh Tiwari I don't want to spoil the fun of a one liner :P

I'll give an incomplete hint to you to start with

Notice that the diagonals divide opposite angles into two equal halves each. Let ABCD be a Parallelogram with the Diagonals bisecting at O. (Diagonals of a //gm bisect each other)

Consider Triangle AOB with Angle A as the acute angle

Angle OAB is 30 degrees and Angle ABO is 60 degrees.

Angle AOB becomes 90 degrees. Similarly do so for Triangles AOD, DOC and COB.

You'll notice that you're done.

Mehul Arora - 5 years ago

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@Mehul Arora I agree with @Rishabh Tiwari 's objection. Plus, it is easy to constuct a parallelogram with one angle as 6 0 60^\circ which isn't a rhombus. The fact that the square of the ratio of the diagonals is 3 plays a role in this problem.

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@A Former Brilliant Member Exactly ! Thanx for commenting!

Rishabh Tiwari - 5 years ago

@Mehul Arora Diagonals of a parallelogram don't bisect opposite angles, its the property of a rhombus! Hence I think u r wrong somewhere in that proof \text{proof} .

Rishabh Tiwari - 5 years ago

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@Rishabh Tiwari @Deeparaj Bhat please comment!

Rishabh Tiwari - 5 years ago

can you prove it \text{can you prove it} ?

Rishabh Tiwari - 5 years ago

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