Can You Turn a Right Triangle Into an Isosceles Triangle?

Begin with a $3$-$4$-$5$ right triangle. By appending another triangle along one of its sides, you can extend it into a single larger triangle that is isosceles; for instance, gluing a copy along the short leg makes a $5$-$5$-$8$ isosceles triangle. Counting that one, in how many distinct ways can you append a triangle to a $3$-$4$-$5$ right triangle to make an isosceles triangle?

The Fiddler, Zach Wissner-Gross, September 6, 2024(original post)

Solution

For the original and the appended triangle to fuse into a single triangle, the shared side must be a cevian of the larger triangle, a segment from a vertex to the opposite side, with the $3$-$4$-$5$ as one of the two pieces it cuts. So a construction is fixed by three choices: which side of the $3$-$4$-$5$ is the cevian, which of its endpoints is the apex of the big triangle, and which two sides of the big triangle are made equal. Running through all $3\times2\times3$ cases and discarding the degenerate ones (where the appended piece shrinks to nothing) leaves seven genuine isosceles triangles, $$\begin{gather*} (\sqrt{10},5,5),\quad (4,4,4\sqrt2),\quad \bigl(4,4,\tfrac{32}{5}\bigr),\quad \bigl(\tfrac{25}{6},\tfrac{25}{6},5\bigr),\\[2pt] (2\sqrt5,5,5),\quad (5,5,6),\quad (5,5,8), \end{gather*}$$ the last being the $5$-$5$-$8$ from the prompt. So there are $$\boxed{7}$$ ways.

The computation

Encode the construction: for each side as cevian and each apex, solve for the point $Z$ along the cevian’s line that makes the big triangle satisfy each of the three isosceles conditions, keep the non-degenerate solutions, and count the distinct triangles.

import numpy as np
def count(a, b):                                   # isosceles triangles from a right triangle, legs a, b
    P = [np.array([0., 0.]), np.array([a, 0.]), np.array([0., b])]
    d = lambda u, v: np.linalg.norm(u - v)
    sols = set()
    for i, j in [(0, 1), (0, 2), (1, 2)]:          # the shared side acts as a cevian
        Y = P[({0, 1, 2} - {i, j}).pop()]
        for X, W in [(P[i], P[j]), (P[j], P[i])]:  # apex X, cevian foot W; Z lies on ray Y->W
            dirn = (W - Y) / np.linalg.norm(W - Y); YW = d(Y, W); XY = d(X, Y)
            for cond in range(3):                  # XY=XZ, XY=YZ, XZ=YZ
                f = lambda t: [XY - d(X, Y + t*dirn), XY - t, d(X, Y + t*dirn) - t][cond]
                ts = np.linspace(YW + 1e-6, 80 * max(a, b), 6000); v = np.array([f(t) for t in ts])
                for ii in np.where(np.diff(np.sign(v)))[0]:
                    lo, hi = ts[ii], ts[ii + 1]
                    for _ in range(60):
                        m = (lo + hi) / 2
                        lo, hi = (m, hi) if np.sign(f(lo)) == np.sign(f(m)) else (lo, m)
                    t = (lo + hi) / 2; Z = Y + t * dirn
                    s = tuple(sorted([round(XY, 3), round(t, 3), round(d(X, Z), 3)]))
                    if t > YW + 1e-3 and s[0] > 1e-2 and s[0] + s[1] > s[2] + 1e-4: sols.add(s)
    return len(sols)
print(count(4., 3.))                               # 7

Extra Credit

For a general right triangle with legs $a$ and $b$, how many such isosceles triangles $N$ can you make? What values can $N$ take?

Solution

Running the same construction for arbitrary legs, the count is remarkably stable: every right triangle gives $$N=\boxed{7},$$ with one exception. When the right triangle is itself isosceles ($a=b$), its mirror symmetry makes several of the seven constructions coincide and others degenerate, and the count collapses to $2$. So the generic answer is $7$, and the right isosceles triangle is the lone special case, giving fewer.

The computation

Reuse the same count routine, sweeping the leg ratio: it returns $7$ for every generic shape and $2$ at $a=b$.

ratios = np.linspace(0.3, 4.0, 60)
print(sorted({count(r, 1.0) for r in ratios if abs(r - 1) > 1e-6}))   # [7]
print(count(1., 1.))                                                  # 2  (isosceles right)