Find Eccentricity Using Polygon Combinations

 Question

Let $p$ be the number of all triangles that can be formed by joining the vertices of a regular polygon $P$ of $n$ sides, and $q$ be the number of all quadrilaterals that can be formed by joining the vertices of $P$.
If

$$p+q=126,$$

then find the eccentricity of the ellipse

$$\frac{x^2}{16}+\frac{y^2}{n}=1.$$

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Question Image

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Short Solution

1. Triangles from $n$ vertices:

   $$p=\left(\genfrac{}{}{0px}{}{n}{3}\right)$$

2. Quadrilaterals from $n$ vertices:

   $$q=\left(\genfrac{}{}{0px}{}{n}{4}\right)$$

3. Given:

   $$\left(\genfrac{}{}{0px}{}{n}{3}\right)+\left(\genfrac{}{}{0px}{}{n}{4}\right)=126$$

4. Solve for $n$.

5. Substitute $n$ as the denominator of $y^{2}$ in the ellipse:

   $$\frac{x^2}{16}+\frac{y^2}{n}=1$$

   Identify $a$ and $b$, then find eccentricity:

   $$e=\sqrt{1-\frac{b^2}{a^{\mathrm{2}}}}$$

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Image Solution

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Conclusion

Step 1: Solve for $n$

$$\left(\genfrac{}{}{0px}{}{n}{3}\right)=\frac{n(n-1)(n-2)}{6},\left(\genfrac{}{}{0px}{}{n}{4}\right)=\frac{n(n-1)(n-2)(n-3)}{\mathrm{24}}$$

So:

$$\frac{n(n-1)(n-2)}{6}+\frac{n(n-1)(n-2)(n-3)}{24}=126$$

Multiply by 24:

$$4n(n-1)(n-2)+n(n-1)(n-2)(n-3)=3024$$

Factor $n(n-1)(n-2)$:

$$n(n-1)(n-2)\big[(n-3)+4\big] = 3024$$
$$n(n-1)(n-2)(n+1) = 3024$$

Check $n=8$:

$$8\cdot 7\cdot 6\cdot 9=3024\text{✓}$$

So, $n = 8$.

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Step 2: Ellipse parameters

Ellipse:

$$\frac{x^2}{16}+\frac{y^2}{8}=1$$

Compare with:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$$

Here:

$$a^2=16,b^2=8$$$$a=4,b=\sqrt{8}=2\sqrt{2}$$

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Step 3: Eccentricity

$$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{\mathrm{2}}}$$

✅ Final Answer:

$$\boxed{e = \dfrac{1}{\sqrt2}}$$

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Video Solution: