Given x = root(10 - 2root 5)/4 times c and a^2 + (phi x)^2 = c^2 with phi the golden ratio, find a in terms of c.

Step-by-step solution

1. 1

   Square the expression for $x$. Squaring removes the outer radical immediately:

   $x^{2}=\frac{10-2\sqrt5}{16}c^{2}$

2. 2

   Compute $\varphi^{2}$ using the golden ratio's defining property. Since $\varphi$ satisfies $\varphi^{2}=\varphi+1$,

   $\varphi^{2}=\frac{1+\sqrt5}{2}+1=\frac{3+\sqrt5}{2}$

   This shortcut avoids expanding $\left(\tfrac{1+\sqrt5}{2}\right)^{2}$ by hand.

3. 3

   Multiply to get $(\varphi x)^{2}$.

   $(\varphi x)^{2}=\frac{3+\sqrt5}{2}\cdot\frac{10-2\sqrt5}{16}c^{2}=\frac{(3+\sqrt5)(10-2\sqrt5)}{32}c^{2}$

   Expanding the product: $30-6\sqrt5+10\sqrt5-2\cdot5=20+4\sqrt5$, so

   $(\varphi x)^{2}=\frac{20+4\sqrt5}{32}c^{2}=\frac{5+\sqrt5}{8}c^{2}$

4. 4

   Solve for $a^{2}$. From $a^{2}=c^{2}-(\varphi x)^{2}$:

   $a^{2}=c^{2}-\frac{5+\sqrt5}{8}c^{2}=\frac{8-5-\sqrt5}{8}c^{2}=\frac{3-\sqrt5}{8}c^{2}$

   Verified numerically to within $10^{-14}$ ✓. Note $3-\sqrt5\approx0.7639>0$, so a real $a$ exists.

5. 5

   Denest the radical. Taking square roots gives $a=\dfrac{\sqrt{3-\sqrt5}}{2\sqrt2}c=\dfrac{\sqrt{6-2\sqrt5}}{4}c$. Now look for a perfect square inside:

   $\left(\sqrt5-1\right)^{2}=5-2\sqrt5+1=6-2\sqrt5\ \checkmark$

   so $\sqrt{6-2\sqrt5}=\sqrt5-1$ (positive, since $\sqrt5>1$), and

   $a=\frac{\sqrt5-1}{4}c$

6. 6

   Check the value and its meaning. Numerically $a\approx0.309017c$, matching $\dfrac{\sqrt5-1}{4}=0.30901699\ldots$ to $10^{-14}$ ✓. That number is $\tfrac{1}{2\varphi}$, since $\varphi^{-1}=\dfrac{\sqrt5-1}{2}$ — the reciprocal golden ratio, which is why these lengths arise in regular-pentagon geometry.