1977 AHSME Problems/Problem 30

Problem

[asy] for (int i=0; i<9; ++i) { draw(dir(10+40*i)--dir(50+40*i)); } draw(dir(50) -- dir(90)); label("$a$", dir(50) -- dir(90), N); draw(dir(10) -- dir(90)); label("$b$", dir(10) -- dir(90), SW); draw(dir(-70) -- dir(90)); label("$d$", dir(-70) -- dir(90), E); //Credit to MSTang for the diagram [/asy]

If $a,b$, and $d$ are the lengths of a side, a shortest diagonal and a longest diagonal, respectively, of a regular nonagon (see adjoining figure), then

$\textbf{(A) }d=a+b\qquad \textbf{(B) }d^2=a^2+b^2\qquad \textbf{(C) }d^2=a^2+ab+b^2\qquad\\ \textbf{(D) }b=\frac{a+d}{2}\qquad  \textbf{(E) }b^2=ad$

Solution 1

By the law of cosines we can get the following expressions for $d^{2}$ and $b^{2}$:

\[d^{2}=2b^{2}(1-\cos (100^\circ))\] \[b^{2}=2a^{2}(1-\cos (140^\circ))\]

We can substitute what we got for $b^2$ into the expression for $d^{2}$:

\[d^{2}=4a^{2}(1-\cos (140^\circ))(1-\cos (100^\circ))=4a^{2}(1-(\cos (100^\circ)+\cos (140^\circ))+\cos (100^\circ)\cos (140^\circ))\]

Now apply sum-to-product and product-to-sum identities: \[d^{2}=4a^{2}(1-(2\cos (120^\circ)\cos (20^\circ))+\frac{1}{2}(\cos (240^\circ)+\cos (40^\circ)))\] Simplifying further gives us: \[d^{2}=4a^{2}(1+\cos (20^\circ)-\frac{1}{4}+\frac{1}{2}\cos (40^\circ))\] After using the fact that $\cos (40^\circ)=2\cos^2 (20^\circ)-1$, it's not hard to see that the expression in the parentheses is equal to $(\cos (20^\circ)+\frac{1}{2})^{2}$. So we can square-root both sides to find the expression for $d$:

\[d=2a(\cos (20^\circ)+\frac{1}{2})\] Now let's look at the expression for $b^2$. We can apply the reverse of the double angle identity to show that $1-\cos (140^\circ)$ equals $2\sin^2 (70^\circ))$. So if we square root the entire expression we get that \[b=2a\sin (70^\circ)=2a\cos (20^\circ)\] We now have everything in terms of $a$. Luckily when we consider choice $\textbf{A}$ we can verify without much work that this must be the answer.

Solution by harita19

  • NOTE: a much easier solution exists by drawing some lines and recognizing that the nonagon is cyclic, but for those of use who use algebra in every geometry problem, this is the best solution.

Solution 2

Using Van Schooten's Theorem, we draw an equilateral triangle in the nonagon to see that $a+b=d$.


Solution 3

Reflect $b$ such that it connects the left vertex of the $a$ to the $2$nd vertex counterclockwise. This length is also $b$. Notice that when connecting the two endpoints of the $b$ segments, we achieve a length of $d$. We now have an isosceles trapezoid with bases $a$ and $d$.

Next, we can find the angle between the $a$ and the $b$ segments as follows. The interior angle of a regular nonagon measures $180-\frac{360}{9}=140^{\circ}$. Thus, the isosceles triangle formed from two of the side lengths of the nonagon and $b$ has angles $140$, $\frac{180-140}{2}$, and $\frac{180-140}{2}=20$. Then, the angle between $a$ and $b$ is $140-20=120$. We can therefore deduce that the other angles are $120$, $60$, and $60$ degrees.

We can drop a perpendicular from both endpoints of segment $a$ to segment $d$ to get a rectangle and two $30$-$60$-$90$ triangles. Each of the triangles has a shorter leg length of $\frac{d-a}{2}$, and hence, we can deduce that $\frac{d-a}{2} \cdot 2=b$. Hence, the correct answer is $\mathbf{(A)} \; d=a+b$

- Solution by scjh999999