Basic Math Problem of the Week 9/14/2017

[PF PotW Robot]

Here is this week's basic math problem. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods.

Given positive real numbers $a,\,b,\,c,\,d$ that satisfy the equalities

$$a^2-ad+d^2=b^2+bc+c^2$$ and $$a^2+b^2=c^2+d^2$$

find all possible values of the expression $\dfrac{ab+cd}{ad+bc}$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[jedishrfu]

Just thinking here. I would start with the Pythagorean triples table where you can find cases of a^2 + b^2 = c^2 + d^2

http://www.tsm-resources.com/alists/trip.html

  3      4      5
  6      8     10     [3 - 4 - 5]
  5     12     13
  9     12     15     [3 - 4 - 5]
  8     15     17
 12     16     20     [3 - 4 - 5]
  7     24     25
 15     20     25     [3 - 4 - 5]
 10     24     26     [5 - 12 - 13]
 20     21     29
 18     24     30     [3 - 4 - 5]
 16     30     34     [8 - 15 - 17]
 21     28     35     [3 - 4 - 5]
 12     35     37
 15     36     39     [5 - 12 - 13]
 24     32     40     [3 - 4 - 5]
  9     40     41
 27     36     45     [3 - 4 - 5]
 14     48     50     [7 - 24 - 25]
 30     40     50     [3 - 4 - 5]
 24     45     51     [8 - 15 - 17]
 20     48     52     [5 - 12 - 13]
 28     45     53
 33     44     55     [3 - 4 - 5]
 40     42     58     [20 - 21 - 29]
 36     48     60     [3 - 4 - 5]
 11     60     61
 16     63     65
 25     60     65     [5 - 12 - 13]
 33     56     65
 39     52     65     [3 - 4 - 5]
 32     60     68     [8 - 15 - 17]
 42     56     70     [3 - 4 - 5]

One such example is 7^2 + 24^2 = 15^2 + 20^2 = 25^2 then check which ones satisfy the first equation and then evaluate the last one to see a pattern.

 

[mfb]

I don't think you'll find many solutions if you only search through a very limited set of integers.

You can fix a and b and then solve for c and d, for example. Add one small trick and the rest is mechanical solving.

 

[Charles Link]

Since $a^2+b^2=c^2+d^2$, the numerator can be rewritten as $ab+cd= \frac{1}{2}[(a+b)^2-(c-d)^2]$. $\\$ Meanwhile, from the first expression they give, the denominator is $ad+bc= (a^2-b^2)-(c^2-d^2)=(a+b)(a-b)-(c+d)(c-d)$. $\\$ If we let $c=d$, the fraction becomes $F= \frac{1}{2}\frac{(a+b)}{(a-b)}$. I would need to look over the fraction carefully, but it appears it may be something like $|F|>\frac{1}{2}$. I need to spend a little more time evaluating it, and in addition, need to try cases where $c \neq d$. $\\$ Editing: Setting $c \rightarrow 0$ , it appears all positive values of the fraction are possible, and setting $d \rightarrow 0$, it appears all negative values of the fraction are possible. The numerator is simply $ab$ in these cases, and the denominator can be evaluated by using $a^2+b^2=d^2$ and $a^2 +b^2=c^2$ for the cases of $c=0$ and $d=0$ respectively. Since $a,b,c, \, and \, d$>0, the numerator can never be equal to zero, so zero is the one and only value that the fraction can not take on.