Difference between revisions of "1998 AIME Problems/Problem 10"

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Thus <math>a + b + c = 100 + 50 + 2 = \boxed{152}</math>.
 
Thus <math>a + b + c = 100 + 50 + 2 = \boxed{152}</math>.
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==Solution 2==
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Isolate a triangle, with base length <math>200</math> (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as <math>x</math>. Since the interior angle is <math>45</math> degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get:
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<cmath>\begin{eqnarray*}
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200^2 &=& 2x^2 - 2x^2*cos(45^\circ) \ &=& 2x^2 - 2x^2*\frac{\sqrt{2}}{2} \ &=& (2-\sqrt{2})x^2
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\end{eqnarray*}</cmath>
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And thus <cmath>x = \frac{200}{\sqrt{2-\sqrt{2}}}</cmath>
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From the above, <math>x = 20\sqrt{r}</math>, so we get
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<cmath>\begin{eqnarray*}
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r &=& (\frac{200}{20(\sqrt{2-\sqrt{2}})})^2 \ &=& (\frac{10}{\sqrt{2-\sqrt{2}})})^2 \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} \ &=& \frac{200 + 100\sqrt{2}}{2} \ &=& 100 + 50\sqrt{2}
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\end{eqnarray*}</cmath>
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And hence the answer is <math>100 + 50 + 2 \Rightarrow \boxed{152}</math>
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== See also ==
 
== See also ==

Latest revision as of 12:02, 5 August 2019

Problem

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a + b + c$.

Solution

The key is to realize the significance that the figures are spheres, not circles. The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.

1998 AIME-10a.png

Let us examine the relation between one of the outside 8 spheres and the center one (with radius $r$):

1998 AIME-10b.png

If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$. Then by the Pythagorean Theorem:

\[x^2 + (r-100)^2 = (r+100)^2 \Longrightarrow x = 20\sqrt{r}\]

1998 AIME-10c.png

$x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\sqrt{r}$. We can draw another right triangle as shown above. One leg has a length of $200$. The other can be found by partitioning the leg into three sections and using $45-45-90 \triangle$s to see that the leg is $100\sqrt{2} + 200 + 100\sqrt{2} = 200(\sqrt{2} + 1)$. Pythagorean Theorem:

\begin{eqnarray*} (40\sqrt{r})^2 &=& 200^2 + [200(\sqrt{2}+1)]^2\\ 1600r &=& 200^2[(1 + \sqrt{2})^2 + 1] \\ r &=& 100 + 50\sqrt{2} \end{eqnarray*}

Thus $a + b + c = 100 + 50 + 2 = \boxed{152}$.

Solution 2

Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$. Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*cos(45^\circ) \\ &=& 2x^2 - 2x^2*\frac{\sqrt{2}}{2} \\ &=& (2-\sqrt{2})x^2 \end{eqnarray*}

And thus \[x = \frac{200}{\sqrt{2-\sqrt{2}}}\]

From the above, $x = 20\sqrt{r}$, so we get

\begin{eqnarray*} r &=& (\frac{200}{20(\sqrt{2-\sqrt{2}})})^2 \\ &=& (\frac{10}{\sqrt{2-\sqrt{2}})})^2 \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ &=& \frac{200 + 100\sqrt{2}}{2} \\ &=& 100 + 50\sqrt{2} \end{eqnarray*}

And hence the answer is $100 + 50 + 2 \Rightarrow \boxed{152}$



See also

1998 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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