# Revival-Math Q&A

#### v8archie

Math Team
Ah! It's that $2^{206}$. It's better as $2^23^{136}$.

So$$3^{670}2^2$$ is the next guess.

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#### Hoempa

Math Team
That's it! Fortunately, my question was a bit more of a challenge than I expected.

#### v8archie

Math Team
I'll have a think and post one tonight.

We have an algorithm then, for any $a$. I think with a bit of thought we can get to the answer too, because the 206 was a function of $a$ and $n$, and $n$ is a function of $a$ and $e$.

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#### Hoempa

Math Team
For the integer terms, there are three cases; the classes mod 3.
For the non-integer classes it's merely
$$\displaystyle \text{max}\left(\left(\frac{a}{\lfloor \frac{a}{e}\rfloor} \right)^{\lfloor \frac{a}{e} \rfloor},\left(\frac{a}{\lceil \frac{a}{e}\rceil} \right)^{\lceil \frac{a}{e} \rceil}\right)$$
For positive integer a.

I might leave some time for trying to answer your next question to give at least Balarka a chance, he much liked to get control I believe.

#### v8archie

Math Team
OK then. Geometry is not my strong suit, so I haven't yet solved this. It's taken from a admission paper for a Japanese university.

Consider a triangle $ABC$ where $\angle A = 60^\circ$. Let $O$ be the inscribed circle of triangle $ABC$, as shown in the figure.

Let $D$, $E$ and $F$ be the points at which the circle $O$ is tangent to the sides $AB$, $BC$ and $CA$. And let $G$ be the point of intersection of the line segment $AE$ and the circle $O$.

1. Let $\triangle{ADF}$ be the area of the triangle ADF. Compute $$\frac{\triangle{ADF}}{AG \cdot AE}$$
2. Set $x = AD$. When $BD=4$ and $CF=2$, what is the length of $BC$? Show that $x$ satisfies a quadratic equation and solve it for $x$.

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#### eddybob123

1. We know that $$\displaystyle AG\cdot AE = AD^2$$ by the tangent-secant theorem. The area of $$\displaystyle \triangle ADF$$ is $$\displaystyle AD\cdot AF \cdot \sin 60 \cdot \frac{1}{2}$$, which is just $$\displaystyle AD^2\cdot\sqrt{3}$$, so the ratio is simply $$\displaystyle \sqrt{3}$$.

I don't have a solution for 2. yet. I'm going to work on it tonight, after the first World Cup Match.

#### v8archie

Math Team
1. We know that $$\displaystyle AG\cdot AE = AD^2$$ by the tangent-secant theorem. The area of $$\displaystyle \triangle ADF$$ is $$\displaystyle AD\cdot AF \cdot \sin 60 \cdot \frac{1}{2}$$, which is just $$\displaystyle AD^2\cdot\sqrt{3}$$, so the ratio is simply $$\displaystyle \sqrt{3}$$.
Thanks for the tangent-secant theorem. Not one I knew.

I rather think you have an error in the area of the triangle though.

#### eddybob123

I rather think you have an error in the area of the triangle though.
Oh, darn. it should be $$\displaystyle \frac{\sqrt{3}}{4}$$

#### v8archie

Math Team
Doesn't the tangent-secant theorem require the secant to be an extended diameter?

#### eddybob123

Doesn't the tangent-secant theorem require the secant to be an extended diameter?
Of course not.

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