Can You Find An Extra Perfect Square?

Riddler Express

Two congruent larger squares each touch a circle at one corner, and a smaller square, tilted $45^\circ$, touches the circle at two of its corners. How many times larger is a big square’s area than the small one’s?

Solution

Let the circle have centre $O$, the big squares side $x$ and the tilted square side $y$. The tilt makes the small square’s diagonal horizontal of length $y\sqrt2$, so its corners stick out $\tfrac{y}{\sqrt2}$ to each side. Reading off the figure, a corner $B$ of a big square sits at horizontal distance $OE = \tfrac12\big(x + \tfrac{y}{\sqrt2}\big)$ from the axis and height $BE = x$, while a corner $A$ of the small square sits at $OF = OE + \tfrac{y}{\sqrt2}$ and $AF = x - \tfrac{y}{\sqrt2}$. Both corners lie on the circle, so they are equidistant from $O$: $$OF^2 + AF^2 = OE^2 + BE^2.$$ Expanding and cancelling collapses neatly to $-\tfrac{x}{\sqrt2} + \tfrac{3y}{2} = 0$, that is $\tfrac{x}{y} = \tfrac{3}{\sqrt2}$, and squaring, $$\frac{x^2}{y^2} = \boxed{\tfrac92}.$$

The computation

Set $y = 1$ and solve the equidistance condition (both corners on one circle) for $x$, then report $x^2/y^2$.

from scipy.optimize import fsolve
from math import sqrt
def equal_radii(x):
    x = x[0]
    OE = 0.5 * (x + 1 / sqrt(2)); BE = x
    AF = x - 1 / sqrt(2); OF = OE + 1 / sqrt(2)
    return [OF**2 + AF**2 - (OE**2 + BE**2)]
x = fsolve(equal_radii, [2.0])[0]
print(x**2)                                    # 4.5 = 9/2

Riddler Classic

For some perfect squares, deleting the last digit leaves a different perfect square (for instance $256 \to 25$). After $16, 49, 169, 256, 361$, what are the next three? How many are there? Extra credit: $169 \to 16 \to 1$ stays square even after deleting two digits; is there another square like that?

Solution

Square each whole number, chop off its last digit, and keep the cases where what remains is itself a perfect square. After the five given, the next three are $$\boxed{1444,\ 3249,\ 18496}$$ ($38^2 \to 144 = 12^2$, $57^2 \to 324 = 18^2$, $136^2 \to 1849 = 43^2$), and the sequence never stops: $64009, 237169, 364816, \dots$ keep coming, so there are infinitely many.

The double property of $169$ is far rarer. Searching every square up to $10^{16}$ for one whose last two digits can both be peeled off to leave squares each time turns up nothing else: $$\boxed{169 \text{ stands alone.}}$$

The computation

Test each square by deleting its last digit and checking whether the result is a perfect square; for the extra credit, require the twice-shortened number to be square as well.

from math import isqrt
def is_sq(m): return m >= 0 and isqrt(m)**2 == m
singles = [n*n for n in range(4, 200) if is_sq((n*n) // 10)]
print(singles[:8])                             # ...1444, 3249, 18496
doubles = [n*n for n in range(4, 2_000_000)
           if is_sq((n*n) // 10) and (n*n) // 100 >= 1 and is_sq((n*n) // 100)]
print(doubles)                                  # [169]