Revival-Math Q&A

v8archie

Math Team
Dec 2013
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Ah! It's that $2^{206}$. It's better as $2^23^{136}$.

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

187629297917542688320603678157614365904524418412591178142726894140044463553152280598824620966838774676790770981761627769101055941652908481919393148196988061079507078293574977646870161972504324686904709749601943235520999089337796555183844156493668213756787443251779936444734558862589413095668201611795605304054798652615396
 
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Hoempa

Math Team
Apr 2010
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That's it! Fortunately, my question was a bit more of a challenge than I expected.
Your turn
 

v8archie

Math Team
Dec 2013
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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
Apr 2010
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361
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
Dec 2013
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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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Sep 2012
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British Columbia, Canada
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. :D
 

v8archie

Math Team
Dec 2013
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Colombia
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.
 

v8archie

Math Team
Dec 2013
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Doesn't the tangent-secant theorem require the secant to be an extended diameter?
 
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