Yes, when I first encountered this one, I spent over 3 minutes trying for a formal solution before seeing that it needs to be greater than 35 ðŸ˜¦ And the other method I noticed only on revisiting the problem later – at the second try it struck me quite quickly ðŸ˜› Very tricky one to visualise.

That was a deadly problem… one of the best u have posted…

Yes, when I first encountered this one, I spent over 3 minutes trying for a formal solution before seeing that it needs to be greater than 35 ðŸ˜¦ And the other method I noticed only on revisiting the problem later – at the second try it struck me quite quickly ðŸ˜› Very tricky one to visualise.

regards

J

Seriously awesome!

(Y)

This is a very good problem . Is there another way to solve this problem ?

Almost certainly, yes. Most problems would have multiple possible ways of solving. Try and find one!

regards

J

what do you think about my solution ?

Here is how I solved this question:

A( ) — Denotes angle

S( ) — Denotes side

Construction : Join BD

Obseration : A( CBD ) = A( CBD )

So,

A( CBD ) = [ 180 – (y+20) ] / 2 = ( 160 – y ) / 2

A( CDB ) = [ 180 – (y+20) ] / 2 = ( 160 – y ) / 2

A( ABD ) = 140 – A( CBD) = (120 + y) / 2

A( ADB ) = 110 – A( CDB ) = ( 60 + y ) / 2

A( BAD ) = x + 20

Assumption : y S(DC) > S(AD)

Also , in triangle ABD we know that A( ABD ) > A( ADB )

(2) Therefore , in triangle ABD ==> S(AD) > S(AB)

But , we know that S(DC) = S(AB) . Hence , (1) and (2) contradicts each other that means our assumption is wrong .

Now checking values of x and y for the following cases:

y x Compare

20 50 y < x

30 40 y x

So, the correct option is (3) .

There was a formatting problem above .

Assumption is y S(DC) >S(AD)

Assumption y is LESS than x . (1) in triangle ADC side DC is GREATER than side AD