The unit-sided square $ABCD$ has its midpoint $F$ on side $AB$. A circle centered at $F$ passing through $C$ intersects the half-line $AB$ at $G$. The circle with diameter $AG$ intersects the segment $BC$ at $H$. Let the point $E$ be the closer quarter point of the diagonal $AC$ to $C$, and the lengths of segments $BE$, $BH$, and the semicircle arc $BF$ be $e$, $h$, and $f$ respectively.

Determine the value of $\frac{e-f}{h-f}$ to two significant figures.

To what number of significant figures must the value of $\pi$ be known for this?
We saw in problem F. 2250 that the number $h=\sqrt{\frac{\sqrt{5}-1}{2}}=0.78615 \ldots$ is a good approximate value for $f=\pi / 4=0.78539 \ldots$ (with an error less than $1 \%$), where both decimal fractions are lower approximations, accurate to 5 significant digits, and we used the first 6 significant digits of $\pi$: 3.14159. From these, we get limits for the denominator of the expression to be examined, containing only 2 significant digits (due to the approximation):

$$
0.00075 < h - f < 0.00077
$$

However, since the problem requires 2 significant digits in the quotient, we already need to look for limits containing at least 3 significant digits in place of (1), i.e., a more precise approximation for $h$. It is not possible to answer the question if we do not decide on the degree of approximation for $h$ and $e$. However, this is up to us.

![](https://cdn.mathpix.com/cropped/2024_05_02_b69ab8326c2c7577d656g-1.jpg?height=640&width=943&top_left_y=637&top_left_x=580)

Let's write: $h=\sqrt{\sqrt{1.25}-0.5}$. From the school function tables, $\sqrt{1.25}=1.118$ to 4 significant digits, and it is not clear whether this is a lower or upper approximation. It is known that in this case $1.25: 1.118=1.11806797 \ldots$ is also an approximation, and their arithmetic mean: $1.11803398 \ldots$ is a better approximation:

$$
\frac{1.118+\frac{1.25}{1.118}}{2}-\sqrt{1.25}=\frac{(\sqrt{1.25}-1.118)^{2}}{2 \cdot 1.118}<\frac{0.0005^{2}}{2}<2 \cdot 10^{-7}
$$

This is also an upper approximation, as the difference is positive (the starting point of 1.118 was a lower approximation).

Accordingly, rounding the arithmetic mean increased by $2 \cdot 10^{-7}$ to 7 decimal places, we get an upper approximation, and a lower one if we decrease by $2 \cdot 10^{-7}$ and round down. We immediately subtract the 0.5 term:

$$
0.6180337 < h^2 < 0.6180342
$$

The lower approximation of the square root of the left side to 3 significant digits is 0.786, so $0.6180342: 0.786=0.7863030 \ldots$ is an upper approximation, and again the arithmetic mean of these is an upper approximation, thus

$$
0.78615 < h < 0.7861515
$$

From $\pi$ using the digits already used so far

$$
-0.7854 < -f < -0.7853975
$$

so we can use this instead of (1):

$$
0.00075 < h - f < 0.000754
$$

In the numerator of the expression to be examined, $e=\sqrt{10} / 4=\sqrt{0.625}$. Here, $\sqrt{10}$ is a frequently used approximation for $\pi$ - albeit a rougher one - so essentially this is our question: how many times smaller is the error of the approximation in problem 2250 compared to the error of $\pi \approx \sqrt{10}$.

Refining the value taken from the tables similarly to the above

$$
0.7905604 < e < 0.7905695
$$

and for $f$ again using 6 significant digits

$$
0.0051694 < e - f < 0.005172
$$

And now

$$
\frac{0.0051694}{0.000754} < \frac{e - f}{h - f} < \frac{0.005172}{0.00075}
$$

Here the lower limit is $6.855 \ldots$, the upper limit is 6.896, so we got that it was enough to know the value of $\pi$ to 6 significant digits, and the quotient to 2 significant digits is 6.9.

With 5 significant digits, 3.1416 (the $3.14155 < \pi < 3.14165$ interval) on the right side of (1a) 0.000754 is replaced by 0.0007765 and on the left side of (2) $6.65 \ldots$ This is not sufficient to enclose the quotient in an interval for which the result of rounding to 2 significant digits is unambiguous for every number in the interval.


[^0]:    ${ }^{1}$ K. M. L. 61 (1980), p. 213.