Regular pyramid and sphere

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Regular pyramid and sphere

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Source: UK IMO NST 2

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392 posts

kueh

#1 h Jun 1, 2005, 12:33 PM • 2 Y

Y by Adventure10, Mango247

Let $n \geq 2$ be a natural number. A pyramid $P$ has base $A_1A_2...A_{2n}$ and apex $O$ . The polygon $A_1A_2...A_{2n}$ is regular and the point $C$ is its centre. The line $OC$ is perpendicular to the plane of the base of $P$ . A sphere passes through $O$ and meets each of the line segments $OA_i$ internally. For each $i = 1, 2,..., 2n$ , let $X_i$ be the point (other than $O$ ) where the sphere meets $OA_i$ . Prove that $OX_1+OX_3 + ... + OX_{2n-1} = OX_2+OX_4 + ... + OX_{2n}$

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#2 h Jun 1, 2005, 5:55 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

We can actually project all the points and segments to the plane of the base and thus get back a plane figure. Now it should be easy to handle.

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#3 h Jun 1, 2005, 7:45 PM • 2 Y

Y by Adventure10, Mango247

Let $U$ be the center of sphere and $V$ it's projection on a plane of poligon.

$A_iX.A_iO = A_iO(A_iO - OX_i)=A_iU^2 - R^2 = A_iV^2 +UV^2 -R^2$

Summing we just need to prove that for any point $V$ in plane $A_1A_2..A_n$ that

$A_1V^2+A_3V^2+...+A_{2n-1}V^2 = A_2V^2+A_4V^2+...+A_{2n}V^2$

But this is corect from analitic geometry and complex numbers.

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#4 h Jun 1, 2005, 9:37 PM • 2 Y

Y by Adventure10, Mango247

Use intersecting chords theorem to make the condition equivalent to some sums of squares being equal, then use the gpat (I missed this last step, but it is still 7-, probably, as it is an easy application, and I proved it for 2 divides n)

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kueh

#5 h Jun 8, 2005, 6:35 PM • 2 Y

Y by Adventure10, Mango247

cadge_nottosh wrote:

Use intersecting chords theorem to make the condition equivalent to some sums of squares being equal, then use the gpat (I missed this last step, but it is still 7-, probably, as it is an easy application, and I proved it for 2 divides n)

interesecting chords thm? where's the cyclic quad?

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#6 h Jun 9, 2005, 8:24 AM • 2 Y

Y by Adventure10, Mango247

Well, it is easy to see that it suffices to prove $\Sigma A_{2i+1}X_{2i+1} = \Sigma A_{2i}X_{2i}$ .
But if $K$ is the centre of the sphere, $R$ its radius, then $A_{i}X_{i}.A_{i}O = A_{i}K^{2} - R^{2}$ .
Then GPAT finishes it off.

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Frozen

#7 h Oct 17, 2005, 4:19 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

What is GPAT?

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#8 h Oct 17, 2005, 5:48 PM • 1 Y

Y by Adventure10

Well, if $(X_{i},M_{i}) , (Y_{i}, N_{i})$ are two sets of points with masses associated with them, with the total mass of $X$ being $M$ , of $Y$ being $N$ , and the mean square distance of the two sets of points is defined to be $\frac{1}{MN}\sum{M_{i}N_{i}|X_{i}Y_{i}|^2 }$ , with $\sigma(X)$ being the mean square distance of the points $X_i$ to their centroid $X^{\prime}$ , then the Generalised Parallel Axis Theorem states that the mean square distance of the $X_{i}$ and $Y_{i}$ is $\sigma(X) + |X^{\prime}Y^{\prime}|^2 + \sigma(Y)$ .

Jack

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