Geez... a hard geometry problem.

[Alan]

I still cannot solve this one. see if any one can

http://img138.imageshack.us/img138/4060/hardproblem.jpg

find AB

[maikel]

Is CE = 1/2 CB and CD = 1/2 CA given? if yes, it is "just" two three times pythagoras. If not, im still thinking.

[Alan]

yes those are given.

and also yes.

if you can get the length of ED, AB = 2*ED. but i don't know if that can be done.

[maikel]

means exactly:

AB^2 = CA^2 + CB^2                                                       (i)
BD^2 = CB^2 + CD^2  <=>  BD^2 = CB^2 + 1/4 CA^2   (ii)
AE^2 = CE^2 + CA^2  <=>  AE^2  = 1/4 CB^2 + CA^2   (iii)

as you see, to solve you need CA and CB but (ii) and (iii) give you these values. Since there are two equations for two unknown variables.

Hope this helps

Maikel

furthermore:

(iii) <=> CA^2 = AE^2 - 1/4 CB^2 => in (ii) BD^2 = CB^2 + 1/4 (AE^2 - 1/4 CB^2)

AE and BD are known.. and so on...



I have to go to bed now.. so im very sorry not to be able to help more. G'night buddy

[Alan]

Thanks for your help.

I think I solved it.

I just figured it is the average of the two known lengths * 4/3

heres a "proof"

http://img215.imageshack.us/img215/8041/hardproblemproof.jpg

[maikel]

If you say so. I think i do not understand what ya doing there. But maybe you're right. Dunno.


Maikel

[tition]

Observations:

1. The solutions to such problems are usually unique (based on statistical knowledge of mathematics teachers).

2. Teachers usually give integral lengths of the legs of their right-angled triangles.

3. AC=4, BC=10 works, because:

4^2+5^2=41,
10^2+2^2=104=4*26

4. AB=sqrt(4^2+10^2)

5. Therefore either the answer is sqrt(116), or the problem is incorrect because it doesn't have a unique solution, which is implied by the wording of the problem.

[QWERTYman]

See A? See B? See the line between them?

I just found AB.

[tition]

http://www.math.ncku.edu.tw/~cfnien/FindX.jpg

[blackcoder41]

i don't get this

BD^2 = CB^2 + CD^2  <=>  BD^2 = CB^2 + 1/4 CA^2   (ii)
AE^2 = CE^2 + CA^2  <=>  AE^2  = 1/4 CB^2 + CA^2   (iii)

why
BD^2 = CB^2 + CD^2
become
 BD^2 = CB^2 + 1/4 CA^2

and the same with (iii)

[maikel]

maikel wrote:

Is CE = 1/2 CB and CD = 1/2 CA given? if yes, it is "just" two three times pythagoras. If not, im still thinking.

Alan wrote:

yes those are given.

Maikel

[blackcoder41]

yeah why

1/4 CA^2

? shouldn't it be

1/2 CA^2

?

pardon me for my weak mathematics skill, i been skipping class before so...

[maikel]

(1/2 CA)^2 = (1/2)^2 CA^2 = 1/4 CA^2

[blackcoder41]

oh sorry about that.. that's much clearer with parenthesis now..