ABCD is a quadrilateral.

AB|=2

BC=16

CD=13

DA=5

Find |DBC|/|ABD|

Results 1 to 15 of 19

- May 22nd 2017, 07:08 AM #1

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 22nd 2017, 07:34 AM #2

- Joined
- Dec 2016
- From
- Earth
- Posts
- 166
- Thanks
- 73

- May 22nd 2017, 10:40 AM #3

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 22nd 2017, 11:48 AM #4

- Joined
- Nov 2010
- Posts
- 3,397
- Thanks
- 1351

## Re: Olympics1

Use Heron's Formula:

$\dfrac{\sqrt{ \dfrac{|BD|+|DC|+|CB|}{2} \left(\dfrac{|BD|+|DC|+|CB|}{2}-|BD|\right)\left(\dfrac{|BD|+|DC|+|CB|}{2}-|DC|\right)\left(\dfrac{|BD|+|DC|+|CB|}{2}-|CB|\right)}}{\sqrt{\dfrac{|AB|+|BD|+|DA|}{2}\left (\dfrac{|AB|+|BD|+|DA|}{2}-|AB|\right)\left(\dfrac{|AB|+|BD|+|DA|}{2}-|BD|\right)\left(\dfrac{|AB|+|BD|+|DA|}{2}-|DA|\right)}}$

Just plug in the lengths you know and see if anything cancels out.

I get:

$\sqrt{\dfrac{29^2-|BD|^2}{7^2-|BD|^2}}$

- May 22nd 2017, 12:57 PM #5

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 22nd 2017, 01:08 PM #6

- Joined
- Nov 2010
- Posts
- 3,397
- Thanks
- 1351

- May 23rd 2017, 09:32 AM #7

- Joined
- Dec 2016
- From
- Earth
- Posts
- 166
- Thanks
- 73

- May 23rd 2017, 10:46 AM #8

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 23rd 2017, 12:24 PM #9

- Joined
- Nov 2010
- Posts
- 3,397
- Thanks
- 1351

- May 23rd 2017, 01:19 PM #10

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 23rd 2017, 01:28 PM #11

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 23rd 2017, 01:38 PM #12

- Joined
- Nov 2010
- Posts
- 3,397
- Thanks
- 1351

- May 23rd 2017, 02:16 PM #13

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

- May 23rd 2017, 03:23 PM #14

- Joined
- Nov 2010
- Posts
- 3,397
- Thanks
- 1351

## Re: Olympics1

I was able to find BD. Use Heron's formula for each triangle. The sum of the areas is equal to the area of the quadrilateral. Use Brahmagupta's formula for the area of the quadrilateral. BD cannot equal 10. That would mean ABD is not a triangle because $2+5 <10$. The answer that I found involves a simple application of algebra to solve.

$\sqrt{ \dfrac{(29+x)(29-x)(x+3)(x-3)}{16} }+\sqrt{\dfrac{(7+x)(7-x)(x+3)(x-3)}{16} }= \sqrt{2080}$

Note: if the quadrilateral is not cyclic, then the area formula would involve trig and be based of the sum of two opposite interior angles (meaning an infinite number of solutions).

- May 23rd 2017, 03:33 PM #15

- Joined
- Oct 2012
- From
- Istanbul
- Posts
- 680
- Thanks
- 4

## Re: Olympics1

DBC/ABD is 10 according to the book. It is not 6. I mistyped the answer.

To apply Brahmagupta we have to know 2 opposite angles. The formula above looks like Brahmagupta except the part with cosinus.