The sum of $4$ numbers equal to $180$ such that the first number over the second number equal to the fourth number over the third number .How to find these numbers ?

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    $\begingroup$ I'm not entirely sure what you mean here. Do you want $a_1 + a_2 + a_3 + a_4 = 180$ and $a_1/a_2 = a_4/a_3$? $\endgroup$ – Joel Jul 10 '13 at 5:21
  • $\begingroup$ If I understand your question correctly, there are several answers. One of the easiest is $$0+x+(180-x)+0=180$$ where $x$ can be any number whatsoever other than $0$ and $180$. Are there any other conditions? $\endgroup$ – Zev Chonoles Jul 10 '13 at 5:21
  • $\begingroup$ @Zev Chonoles every number is different from others. $\endgroup$ – user85695 Jul 10 '13 at 5:27
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – Gamma Function Jul 10 '13 at 5:27
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    $\begingroup$ Easy. For example use ratio $2$. Then we get $3(c+d)=180$, so $c+d=60$. Now we can pick $c=1, d=59$ or $c=3, d=57$ or (bunch of others). And we can explore various other ratios, like $3$ to $1$, $4$ to $1$, $3$ to $2$, and so on. We will get lots of solutions. $\endgroup$ – André Nicolas Jul 10 '13 at 6:51

Here is an extended hint as to a systematic approach, without all the details filled in.

Note that $\cfrac{a_1}{a_2}=\cfrac{a_4}{a_3}$ implies that both are also equal to $\cfrac{a_1+a_4}{a_2+a_3}$

Now split $180=a+b$ and let $r$ be the highest common factor of $a=rc$ and $b=rd$.

Split $r=p+q$ and allocate $p$ to the first fraction and $q$ to the second fraction.

I've left some things you'll have to think a bit about, because your question shows no work or insight. Like are all the solutions you get by this method different?

Note - this works because $\cfrac {pc}{pd}=\cfrac {qc}{qd}$

  • $\begingroup$ Push it a little further if you don't mind . $\endgroup$ – user85695 Jul 10 '13 at 6:34
  • $\begingroup$ @user85695 If you could show some work or understanding rather than just asking other people to do it all you would get a better response. $\endgroup$ – Mark Bennet Jul 10 '13 at 7:22
  • $\begingroup$ I see ,that's why your answer sound fruitless and vague . $\endgroup$ – user85695 Jul 10 '13 at 11:25
  • $\begingroup$ @user85695 $180=a+b=rc+rd=(p+q)c+(p+q)d=pc+pd+qd+qc$. All you need to do is follow the logic through. Read what Zev Chonoles has put in his comment on your initial question. Just about everyone who has commented on your question could produce a long list of solutions and explain how they formed their list. $\endgroup$ – Mark Bennet Jul 10 '13 at 12:33

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