Notice that AC splits the quadrilateral into two pieces.
ABC is a triangle as well as ADC.
By the Triangle Inequality for triangle ABC, AC must be greater than 6-3=3 and less than 6+3=9. The values are between 4 and 8, inclusive.
By the Triangle Inequaliyt for triangle ADC, AC must be greater than 4-4=0 and less than 4+4=8. The values are between 1 and 7, inclusive.
Thus, the values in the intersection between the two inequalities are from 4-7 for 7-4+1=4 values that satisfy AC, namely 4, 5, 6, and 7 units.
1/a + 1/b = 2/17
Multiply the equation by two...
If 289(17^2) is added to both sides, then (2a-17)(2b-17)=289.
289=1*289 and 17*17.
2a=18 a=9 2b-17=289 2b=153 (9,153), (153,9)
2a-17=17 2a=34 a=17 2b-17=17, 2b=34, b=17 (17,17)
Thus, there should be three(3) positive ordered pairs.
A median bisects the side in half(1/2).
Let the side AC=2a and CB=2B, thus, the medians divide the sides into a and b, respectively.
Systems of equations gives us...
Solve for 2a, and 2b 2\sqrt 85 and b=2\sqrt21
c^2=424 ...solve for c
probably made a mistake somewheree !