Ptolemy's Riddle II

Since $ABCD$ is a cyclic quadrilateral, $\angle CDA + \angle CBA = 180^\circ$ . And since $ABC$ is an equilateral triangle, $\angle CBA = 60^\circ$ . Then $\angle CDA = 120^\circ$ ).

By using cosine rule, $AC^2 = 13^2 = 169 = CD^2 + DA^2 -2(CD)(DA)(\cos 120^\circ) = CD^2 + DA^2 + (CD)(DA)$

Let $CD = x$ and $DA = y$ for some integers $x,y$ . We can then set up a quadratic equation:

$x^2 + xy + (y^2 - 169) = 0$

The quadratic residue must then be perfect square: $y^2 -4(y^2 - 169) = n^2$ for some integer $n$ .

Thus, $26^2 - 3y^2 = n^2$ .

$26^2 - n^2 = 3y^2$

$(26-n)(26+n) = 3y^2$

$26 - n >0$ . $n < 26$

If $26 - n \equiv 1 \pmod 3$ , then $26 + n \equiv 0\pmod 3$ .

If $26 - n \equiv 0 \pmod 3$ , then $26 + n \equiv 1\pmod 3$ .

Thus, $n$ is not a multiple of $3$ .

If $n = 23$ , then $y = 7$ . Then $x = 8$ .

Then by Ptolemy's Theorem , $DB\times 13 = 13\times 7 + 13\times 8$ .

Thus, $DB = CD + DA = 7+8 = 15$ .

As a result, $DC + DB + DA = \boxed{30}$ .

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Let the red lengths be a, b and c as shown. 13 < b < diameter = $\frac{2\times 13}{\sqrt{3}} = 15.011$ and it has to be an integer. So it must be 14 or 15. Just two possibilities! Cheap enough for trial and error :-)

Moreover, $\angle ADB = \angle CDB = 60°$ Hence, $\cos ADB = \frac{1}{2} = \frac{a^2+b^2-13^3}{2ab}$ Try b = 14 and 15 only b=15 gives integer values of a = 7 or 8. So if a = 7, c must be 8 or vice versa. But irrespective of that the sum of the three red lengths is 7 + 8 + 15 = 30.

In face, Ptolemy could have been more impressive by saying, "And there is one more point that does it!" :-) (of course, that follows from symmetry too.)

Let DA=a and DC=b

Angle ADC=120*

From cosine rule,

$a^2 + b^2 +ab = 13^2 =169$

$(a+b)^2 - ab = 225 - 56$

$(a+b)^2 - ab = (7+8)^2 - (7*8)$

On comparing,

a=7,b=8

From Ptolmey's theorem,

13a + 13b = 13.BD

Therefore, a+b = BD

a+b+BD = 2(a+b)

=2(7+8)

=30

1- realize that DA=DC from equal angles of 60° facing same arcs respectively

2- apply Ptolemy's Theorem so that DB=DC+DA= 2DA

3- apply cosine law in ACD triangle to get that DA = 13/SQRT(3)

4- Sum = 4 DA = 52/sqrt(3) = 30