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## Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

## Distance Time – AIME 2012

When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile.

• is 107
• is 279
• is 840
• cannot be determined from the given information

### Key Concepts

Time

Distance

Speed

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

## Try with Hints

After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Then
$\frac{a}{6} + \frac{n-a}{4}$ = $\frac{n-a}{6} + \frac{2a}{5}$ implies that $a = \frac{5}{19}n$

Then integral value of n is 19 and a = 5 and $t = \frac{13}{3}$ hours that is 260 minutes. Then $19 + 260 = {279}$.

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## Planes and distance | AIME I, 2011 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and Distance.

## Planes and distance- AIME I, 2011

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as $\frac{r-s^\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \lt 1000$, find r+s+t.

• is 107
• is 330
• is 840
• cannot be determined from the given information

### Key Concepts

Plane

Distance

Algebra

AIME I, 2011, Question 13

Geometry Revisited by Coxeter

## Try with Hints

Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d

$\frac{a10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}$=10-d and $\frac{b10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}$=11-d and $\frac{c10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}$=12-d

squaring and adding $100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}$ then having 11-d=y, 100=3$y^{2}$+2then y=$\frac{98}{3}^\frac{1}{2}$ then d=11-$\frac{98}{3}^\frac{1}{2}$=$\frac{33-294^\frac{1}{2}}{3}$ then 33+294+3=330.

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## Rolling ball Problem | Semicircle |AMC 8- 2013 -|Problem 25

Try this beautiful problem from Geometry based on a rolling ball on a semicircular track.

## A Ball rolling Problem from AMC-8, 2013

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3  semicircular arcs whose radii are $R_1=100$ inches ,$R_2=60$ inches ,and $R_3=80$ inches respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from  A to B?

• $235 \pi$
• $238\pi$
• $240 \pi$

### Key Concepts

Geometry

circumference of a semicircle

Circle

Answer:$238 \pi$

AMC-8, 2013 problem 25

Pre College Mathematics

## Try with Hints

Find the circumference of semicircle….

Can you now finish the problem ……….

Find the total distance by the ball….

can you finish the problem……..

The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses $2\pi \times \frac{2}{2}=2\pi$  inches each, and it gains $2\pi$ inches on B .

So, the departure from the length of the track means that the answer is

$\frac{200+120+160}{2} \times \pi$ + (-2-2+2) $\times \pi$=240$\pi$ -2$\pi$=238$\pi$