Categories
Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 279.

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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Categories
AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 330.

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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Categories
AMC 8 Geometry Math Olympiad

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?

Path of the rolling ball- Problem

  • \( 235 \pi\)
  • \( 238\pi\)
  • \( 240 \pi\)

Key Concepts


Geometry

circumference of a semicircle

Circle

Check the Answer


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……..

rolling ball 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\(\pi\)

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