## shubhamsrg 3 years ago now the question which follows up is : If all the sides and area of a triangle were rational numbers then show that the triangle is got by `pasting' two right-angled triangles having the same property

1. anonymous

same property means?

2. shubhamsrg

both triangles should have rational sides.

3. shubhamsrg

|dw:1345129143913:dw| if ABC has rational sides, we have to show AMB and MBC also has rational sides.

4. anonymous

AC.MB is rational because area is rational and AC is rational so MB is rational ...now u just proveAM and MC are rational

5. anonymous

lol...not that easy

6. anonymous

I think it will be enough to show just one side Am or MC rational

7. anonymous

yeah..

8. anonymous

beause if one is rational than the other must be rational

9. shubhamsrg

hmm..

10. shubhamsrg

if we say tanA = BC/AC = rational also tanA = MB/AM is it good enough to say here that AM will be rational ?

11. shubhamsrg

sorry,,tanA=BC/Ab ,,but still rational..

12. anonymous

ABC is right triangle !!? work on $\sqrt{AB^2-MB^2}+\sqrt{BC^2-MB^2}=AM+MC$

13. anonymous

i thought its not right

14. shubhamsrg

hmm,,well,,how will sum of sides help ?

15. anonymous

sorry that will not help..

16. anonymous

AM^2 and MC^2 will be rational

17. anonymous

right?

18. anonymous

right

19. shubhamsrg

yep..thats where even i was wondering

20. shubhamsrg

this ques is a subpart of my previous ques,,do both have any relation ?

21. shubhamsrg

how ?

22. anonymous

i mean ABC is a right triangle?

23. anonymous

Dont think so

24. shubhamsrg

aha...thats what i had assumed!! so its not a right triangle!! hmm..

25. anonymous

sauravshakya mentiond a good point $(AM+MC)^2=AC^2=AM^2+2AM.MC+MC^2$so $$AM.MC$$ is rational

26. anonymous

now our problem is like this if $$a+b$$ and $$ab$$ are rational prove that both $$a$$ and $$b$$ are rational

27. anonymous

YEP

28. shubhamsrg

well why were AM^2 and MC^2 rational ?

29. anonymous

Take right triangle AMB and triangle BMC

30. shubhamsrg

ohh leave it got it..

31. shubhamsrg

if ab and and a+b are rational,,its not necessary both are rational,, eg. -sqrt(2) +1 and sqrt(2) + 1

32. shubhamsrg

has to be an other way around..

33. anonymous

yeah i was thinkin of that

34. shubhamsrg

hmm..

35. anonymous

guys i gotta go...but i love to know what is the answer

36. shubhamsrg

|dw:1345131550151:dw| here,,we have AC/sin(a+b) = BC/sinx now sin(a+b) = sina cosb + cosa sinb = AM/AB . BM/BC + BM/AB . MC/BC = (AM+BM)/AB + (BM +MC)/BC also, sin x =BM/AB on substituing, AC . BM/AB = BC . ( (AM+BM)/AB + (BM +MC)/BC ) AC.BM = (AM+BM)BC + (MB +MC)AB AC BM = AM BC + BM BC + MB AB + MC AB I DONT KNOW IF IT HELPS OR NOT!! lol.. :P

37. experimentX

hasn't this been solved yet?

38. shubhamsrg

its a diff ques..

39. experimentX

i lost track of it ... it's fuzzy ... well what's the Q?

40. anonymous

|dw:1345132117021:dw|if ABC has rational sides, we have to show AMB and MBC also has rational sides.

41. anonymous

@experimentX thats the question

42. experimentX

|dw:1345138702333:dw| a^2 + b^2 = c^2 b^2 = rational AM + MC = rational AM^2 + MC^2 = rational. AM*MC = rational.

43. experimentX

Let me finish it man!! AM + MC = rational <-- either both are irrational or rational) AM*MC = rational <--- either both are irrational or rational Let's assume both to be rational, AM ( rational - AM) = rational AM (rational) - AM^2 = rational irrational - rational = rational (contradiction) hence both must be rational

44. experimentX

typo:- Let's assume both to be irrational,

45. anonymous

here is my proof|dw:1345140385316:dw|

46. anonymous

$$x+y=c$$ is rational Area=$$ch/2$$=rational so $$h$$ is rational since $$c$$ is rational we have$x^2+h^2=a^2$$y^2+h^2=b^2$$(x-y)(x+y)=(x-y).c=a^2-b^2$so we obtain$x-y=\frac{a^2-b^2}{c}$$x+y=c$it gives$x=\frac{a^2+c^2-b^2}{2c}$$y=\frac{b^2+c^2-a^2}{2c}$$$x$$ and $$y$$ both are rational

47. experimentX

lol ... your method is more obvious!!

48. shubhamsrg

aha..i see and understand both methods! thanks again @mukushla and @experimentX