$ABCD$ is a square with $AB=13$ . Points $E$ and $F$ are exterior to $ABCD$ such that $BE=DF=5$ and $AE=CF=12$ .

If the length of $EF$ can be represented as $a\sqrt b,$ where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime, then find $ab$ .

The answer is 34.

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It isn't immediately obvious that adding G and H like that would have FCH and FDG be straight lines.

May be better to explicitly state 5²+12²=13², therefore CFD is a right angle; and then the angles at C are equal to the angles in a triangle = 180, therefore FCH is a straight line.

Or create G and H by extending the lines we're given; then prove using the angles that CDF and BCH are identical so you can fill in the sides.

Angel Wedge
- 4 years, 1 month ago

This is the same diagram as used in the Chinese proof of the Pythagorean theorem.

Much more elegant than my brute force solution....

Joe Horton
- 2 years, 11 months ago

Nice move to extend the figure to get the right-triangles. That simplifies it easily

Archana Bhisikar
- 1 year, 7 months ago

I still can not see how you get answer 34. According to me answer is 17*sqare root 2 which is 24 to nearest integer.

Anil Shah
- 10 months, 1 week ago

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Now I see my mistake.

Anil Shah
- 10 months, 1 week ago

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This was the solution which I was looking for. Good job.

mietantei conan
- 7 years ago

And I have this whole length of trigometry that ends with $17\sqrt { 2 }$

Julian Poon
- 7 years ago

Not clear,Could you draw afigure showing G & H to complete square EGFH ,Thanks K.K.GARG,India

Krishna Garg
- 6 years, 12 months ago

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How did you calculate the coordinates of F and E ? PLEASE ELABORATE.

Manvendra Singh
- 5 years, 6 months ago

I did the same.

Eloy Machado
- 4 years, 8 months ago

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Extend line $\overline{DF}$ to the lower left by $5 units$ . Call the end of the extension as point $P$

Making $\triangle{DPB}$ with sides $\overline{PB}$ parallel and equal to $\overline{FE}$

$\overline{DP} = 10$

$\overline{DB} = 13\sqrt{2}$

$\overline{PB} = \overline{FE}$

$\angle{BDP} = 45 + \cos^{-1} (\frac{5}{13})$

take cosine law

$(\overline{FE})^2 = (\overline{PB})^2 = 10^2 + (13\sqrt{2})^2 - 2(10)(13\sqrt{2})\cos [45 + \cos^{-1} (\frac{5}{13})]$

$\overline{FE} = 17\sqrt{2}$

$ab = 17 \times 2 = \boxed {34}$

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EF² = (13 + 120/13)² + (2
*
(25/13) - 13)²=578 or EF = 17√2. Hence, ab = 17
*
(2 )= 34

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Just introduce the co-ordinate system with origin anywhere........... Find the co-ordinates of E and F by using basic trigonometry and use distance formula to find the length EF

Aditya Anilkumar
- 7 years ago

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Exactly! This is what I did. Actually only the distance formula without any trigonometry will do the job. You can take the lower left vertex of the square as the origin.

Abhinav Chauhan
- 7 years ago

Please explain how did you get 120/13 and 25/13 figures to get EF square values? Thanks K.K.GARG,India

Krishna Garg
- 7 years ago

If lines parallel to BC & DC re drawn resply. thru E and F intersecting in G then from basic trigo.in rt. triangle EFG, EG = 130+120/13 while FG =13 - 2(25/13) So EF² = (13 + 120/13)² + (13 -2(25/13))²

Ajit Athle
- 7 years ago

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Solution

I added Point G and H which is the same as E and F .

Point EF - Hypotenuse

FH=HE=17

45-45-90 triangle

EF= $7\sqrt { 2 }$

17(2) = $\boxed{34}$

Sorry for the bad edited image.