Back To Back Right Triangles

In general, if the side lengths of an integer-length, right-angled triangle are $x$ and $12$ and the hypotenuse has length $y$ then

$x^{2} + 12^{2} = y^{2} \Longrightarrow 144 = y^{2} - x^{2} \Longrightarrow 144 = (y - x)(y + x)$ .

Now as $144 = 2^{4}*3^{2}$ and since $0 \lt y - x \lt y + x$ we have the following cases to consider:

• $y - x = 1, y + x = 144 \Longrightarrow (y - x) + (y + x) = 145 \Longrightarrow 2y = 145$ , leaving $y$ non-integral,

• $y - x = 2, y + x = 72 \Longrightarrow 2y = 74 \Longrightarrow y = 37, x = 35$ ,

• $y - x = 3, y + x = 48 \Longrightarrow 2y = 51$ , leaving $y$ non-integral,

• $y - x = 4, y - x = 36 \Longrightarrow 2y = 40 \Longrightarrow y = 20, x = 16$ ,

• $y - x = 6, y + x = 24 \Longrightarrow 2y = 30 \Longrightarrow y = 15, x = 9$ ,

• $y - x = 8, y + x = 18 \Longrightarrow 2y = 26 \Longrightarrow y = 13, x = 5$ ,

• $y - x = 9, y + x = 16 \Longrightarrow 2y = 25$ , leaving $y$ non-integral.

So the possible lengths for the remaining side $x$ are $5,9,16$ and $35$ . The possible values for $|BC| = |BD| + |DC|$ are then $14, 21, 25, 40, 44$ and $51$ . These values correspond to respective side length values for $\Delta ABC$ of

$(13,14,15), (13,20,21), (15,20,25), (13,37,40), (15,37,44), (20,37,51)$ .

The only one of these triples which represents a right triangle is $(15,20,25)$ , and so $|BC| = \boxed{25}$ is the correct option.

There are many Pythagorean triples with a side length of 12, namely $9-12-15, 12-16-20, 5-12-13, 12-35-37$ . So, it may seem that we don't have much control over what BD and DC are.

However, note that we also require $ABC$ to be a right triangle, and quick angle chasing shows that $BAC, BDA,$ and $ADC$ are similar, which means that they are multiples of the same primitive Pythagorean triangle. Since $AD$ in these triangles correspond to different bases of the primitive Pythagorean triangle, this means that the bases must be a factor of 12, of which the only such triangle is $3-4-5$ .

Hence, triangles $BDA$ and $ADC$ must be $9-12-15, 12-16-20$ (in either order). This gives us $BC = 9 + 16 = 25$ .

Let PPT mean Primitive Pythagorean Triple. BD and DC are not equal.
So 12 must have at least two distinct factors, both of them part of PPT.

First factor of 12 will be used by one leg of PPT where the remaining will be the multiplier. For the second PPT factor also, the remaining will be the multiplier.
3 and 4 is the only chose..
So for 3 as first leg of PPT 3-4-5 in triangle ADC, multiplier is 4. The other leg is 4 . That is BD=4 * 4= 16 .
And for 4 as leg of PPT 3-4-5 in triangle ADB, multiplier is 3. The other leg is 3 . That is DC=3 * 3= 9 .
So BC=16+9=25. ... BD and DC can interchange.

The problem could have been made more interesting by making AD=K * 12, where K is any +tive integer and giving range for BC. It will increase interest if AD have more than two factors that are part of PPT.