Euclid's Riddle II

Suppose that $FD = x$ , $AF = xy$ , $BE = b$ and $CF= c$ for positive integers $x,y,b,c$ . The first set of conditions tell us that $b+c \;= \; 2xy$ Then we know that $DG = x(y+1)$ and $CG = xy+c$ , so that $xy$ and $c$ are coprime and Pythagoras' Theorem applied to $CFG$ tells us that $yc \; = \; 2x(y+1)$ Finally using similar triangles, we have $\frac{b}{2x(y+1)-AE} \; = \; \frac{c}{x(y+2)}$ so that $b(y+2) \; < \; 2c(y+1)$

Since $y$ and $y+1$ are coprime, $y$ must divide $2x$ . Since $x$ and $c$ are coprime, $c$ must divide $2(y+1)$ . There are two cases to consider:

• If $c=y+1$ , then $y=2x$ and $b=y^2-y-1$ . The inequality reads $(y^2-y-1)(y+2) < 2(y+1)^2$ , and the only positive integer solutions of this inequality are $y=1,2,3$ . Since $y=2x$ , we must have $y=2$ . But this implies that $b = 1 < 3 = c$ , which is impossible.
• If $c=2(y+1)$ , then $y=x$ and $b=2(y^2-y-1)$ .The inequality again tells us that $(y^2-y-1)(y+2) < 2(y+1)^2$ , and so $y=1,2,3$ are the only possibilities. The case $y=1$ gives $b=-2 < 0$ . The case $y=2$ gives $b = 2 < 6 = c$ . Thus we must have $y=3$ , and so $x=3$ , $b=10$ , $c=8$ .

Thus the area $ABCD$ is $\tfrac12 \times b \times 2x(y+1) - \tfrac12 \times c \times x(y+1) \; = \; \tfrac12(2b-c)x(y+1) \; = \; \boxed{72}$

Considering the right triangle $CFG$ , it has all integer side lengths, making them Pythagorean triples. Then suppose $CF = a$ ; $FG = b$ ; and $GC = c$ for some integers $a,b,c$ .

From the first part, we know that $AF = CG - FC = c-a$ .

Let then $BE = h$ , and $h + a = 2(c-a)$ . $h = 2c - 3a$ .

We also know that $h > a$ , so $2c - 3a > a$ . Hence, $c > 2a$ .

The upper bound of $h$ happens when we create the triangle $AHG$ to $CFG$ with $\angle HAG$ as a right angle.

Then with similarity ratio, $AH = h_{max} = \dfrac{a}{b}(AH) = \dfrac{a(c-a+b)}{b} > h$ .

Thus, $\dfrac{a(c-a+b)}{b} > 2c - 3a$ .

$ac - a^2 > 2cb - 4ab$

$4ab - a^2 > c(2b-a)$

$\dfrac{4ab - a^2}{2b-a} > c$

$2a + \dfrac{a^2}{2b-a} = a(2 + \dfrac{a}{2b-a}) = a(2 + \dfrac{1}{2\frac{b}{a}-1}) > c$

Since $b > a$ , then $\dfrac{b}{a} >1$ .

$\dfrac{2b}{a} - 1 >1$ .

Thus, $\dfrac{1}{2\frac{b}{a}-1} < 1$ , and $a(2 + \dfrac{1}{2\frac{b}{a}-1}) < 3a$ .

Therefore, $3a > a(2 + \dfrac{1}{2\frac{b}{a}-1}) > c > 2a$ . In other words, $3 > \dfrac{c}{a} > 2$ .

Now from $FD | AF$ , we can rewrite as: $\dfrac{a+b-c}{2} | c-a$ . According to Euclid's formula, we can parametize $c= m^2 + n^2$ ; $a = 2mn$ or $b = m^2 - n^2$ or vice versa, depending on which one is even for some co-prime integers $m>n$ .

Then $\dfrac{m^2-n^2+2mn - (m^2+n^2)}{2} | c-a$ .

$mn - n^2 | c-a$ .

$n(m-n) | c-a$ .

For odd $a = m^2 - n^2$ , $c-a = 2n^2$ . Then $m-n | 2n$ . Since $m-n$ is odd, then $m-n| n$ .

Hence, $k(m-n) = n$ for some integer $k$ . $km = (k+1)n$ . However, $gcd(k, k+1) = 1$ for any $k$ , so $k|n$ . And since $gcd(m, n) = 1$ , then $n|k$ .

That means $n = k$ and $m = k+1 = n+1$ .

Returning to the inequality $3 > \dfrac{c}{a} > 2$ , we can substitute in terms of $m,n$ :

$3 > \dfrac{m^2 + n^2}{m^2 - n^2} > 2$ . Then $3 > \dfrac{(n+1)^2 + n^2}{(n+1)^2 - n^2} > 2$ .

$3 > \dfrac{2n^2 + 2n +1}{2n+1} > 2$ .

$2 > \dfrac{2n^2}{2n+1} >1$

Solving for each side, $2n^2 > 2n+1$ . $2n^2 -2n-1>0$ . Solving the quadratic formula, we will obtain $n > \dfrac{1+\sqrt{3}}{2} \approx 1.37$ or $n \geq 2$ .

On the other hand, $2(2n+1) > 2n^2$ . $0 > n^2 - 2n -1$ . Solving the quadratic formula, we will obtain $n < 1+\sqrt{2} \approx 2.41$ or $n \leq 2$ .

The only possible value is $n= 2$ and, thus, $m = 3$ , and Pythagorean triple of $(5, 12,13)$ . However, with this, $h = 2c- 3a = 26 - 15 = 11$ , but $11 > h_{max} = \dfrac{a(c-a+b)}{b} = \dfrac{5(13-5+12)}{12} \approx 8.33$ , which is not applicable.

That means $a = 2mn$ is even, and so $n(m-n)| c-a = (m^2+n^2) - 2mn = (m-n)^2$ . Hence, $n| m-n$ and $n|m$ .

The only applicable value is $n=1$ .

Then similarly for the inequality $3 > \dfrac{c}{a} > 2$ , we can substitute in terms of $m,n$ :

$3 > \dfrac{m^2 + 1}{2m} > 2$ .

Solving for each side, $m^2 +1 > 4m$ . $m^2 -4m+1>0$ . Solving the quadratic formula, we will obtain $m > 2+\sqrt{3} \approx 3.73$ or $m \geq 4$ .

On the other hand, $m^2 +1 < 6m$ . $m^2 -6m+1 < 0$ . Solving the quadratic formula, we will obtain $m < 3+ 2\sqrt{2} \approx 5.83$ or $m \leq 5$ .

Only $m = 4,5$ works, but since $n=1$ is odd, then $m$ must be even and equals to $4$ .

That leads to the Pythagorean triple of $(8, 15,17)$ . $h = 2c- 3a = 34 - 24 = 10$ .

$AF = 9$ . $FD = \dfrac{9+15}{2} - 9 = 3$ , and $3|9$ . $AG = 24$ , and $DG =12$ .

Finally, the area of the quadrilateral $ABCD = \dfrac{10}{2}(24) - \dfrac{8}{2}(12) = 120 - 48 = \boxed{72}$ .