Difference between revisions of "1977 AHSME Problems/Problem 30"
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Now apply sum-to-product and product-to-sum identities: | Now apply sum-to-product and product-to-sum identities: | ||
| − | <cmath>d^{2}=4a^{2}(1-(2\cos (120^\circ)\cos ( | + | <cmath>d^{2}=4a^{2}(1-(2\cos (120^\circ)\cos (20^\circ))+\frac{1}{2}(\cos (240^\circ)+\cos (40^\circ)))</cmath> |
Simplifying further gives us: <cmath>d^{2}=4a^{2}(1+\cos (20^\circ)-\frac{1}{4}+\frac{1}{2}\cos (40^\circ))</cmath> | Simplifying further gives us: <cmath>d^{2}=4a^{2}(1+\cos (20^\circ)-\frac{1}{4}+\frac{1}{2}\cos (40^\circ))</cmath> | ||
After using the fact that <math>\cos (40^\circ)=2\cos^2 (20^\circ)-1</math>, it's not hard to see that the expression in the parentheses is equal to <math>(\cos (20^\circ)+\frac{1}{2})^{2}</math>. So we can square-root both sides to find the expression for <math>d</math>: | After using the fact that <math>\cos (40^\circ)=2\cos^2 (20^\circ)-1</math>, it's not hard to see that the expression in the parentheses is equal to <math>(\cos (20^\circ)+\frac{1}{2})^{2}</math>. So we can square-root both sides to find the expression for <math>d</math>: | ||
Latest revision as of 21:11, 1 November 2024
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]](http://latex.artofproblemsolving.com/c/5/d/c5d372c564b6b059dc5b124a1c926be386b9ff26.png)
If 
, and 
are the lengths of a side, a shortest diagonal and a longest diagonal, respectively,
of a regular nonagon (see adjoining figure), then
Solution 1
By the law of cosines we can get the following expressions for 
and 
:
![\[d^{2}=2b^{2}(1-\cos (100^\circ))\]](http://latex.artofproblemsolving.com/0/4/c/04ccb0543eabcd1eaf705a4eb9f4cfcc2bf92c6a.png)
![\[b^{2}=2a^{2}(1-\cos (140^\circ))\]](http://latex.artofproblemsolving.com/2/4/9/24934de07dfa7a4b8f9cd48546b0a70ae6bfc8e9.png)
We can substitute what we got for 
into the expression for :
![\[d^{2}=4a^{2}(1-\cos (140^\circ))(1-\cos (100^\circ))=4a^{2}(1-(\cos (100^\circ)+\cos (140^\circ))+\cos (100^\circ)\cos (140^\circ))\]](http://latex.artofproblemsolving.com/3/0/9/30917f6297ddf283002ff8f629d5555dac1687b2.png)
Now apply sum-to-product and product-to-sum identities:
![\[d^{2}=4a^{2}(1-(2\cos (120^\circ)\cos (20^\circ))+\frac{1}{2}(\cos (240^\circ)+\cos (40^\circ)))\]](http://latex.artofproblemsolving.com/0/0/7/00765a1db8736058079e8dfb23c6e4afdc7ba5a3.png)
Simplifying further gives us: ![\[d^{2}=4a^{2}(1+\cos (20^\circ)-\frac{1}{4}+\frac{1}{2}\cos (40^\circ))\]](http://latex.artofproblemsolving.com/f/a/8/fa8fc5f7329ccb140ce2360be739c471eceb20e1.png)
After using the fact that 
, it's not hard to see that the expression in the parentheses is equal to 
. So we can square-root both sides to find the expression for :
![\[d=2a(\cos (20^\circ)+\frac{1}{2})\]](http://latex.artofproblemsolving.com/d/c/0/dc09b194d5fb3eec408d4ab98c5e59c6dc33b07b.png)
Now let's look at the expression for . We can apply the reverse of the double angle identity to show that 
equals 
. So if we square root the entire expression we get that ![\[b=2a\sin (70^\circ)=2a\cos (20^\circ)\]](http://latex.artofproblemsolving.com/a/7/f/a7f1e561f968bf57f681f84200d9296d27ebb266.png)
We now have everything in terms of 
. Luckily when we consider choice 
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, and immediately see that 
.
