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 \(19 + 260 = {279}\).
Try this beautiful problem from the Pre-RMO, 2019 based on Covex Cyclic Quadrilateral.
Covex Cyclic Quadrilateral - PRMO 2019
Let ABCD be a convex cyclic quadrilateral. Suppose P is a point in the plane of the quadrilateral such that the sum of its distance from the vertices of ABCD is the least. If {PA,PB,PC,PD}={3,4,6,8}, find the maximum possible area of ABCD.
is 107
is 55
is 840
cannot be determined from the given information
Key Concepts
Largest Area
Quadrilateral
Distance
Check the Answer
Answer: is 55.
PRMO, 2019, Question 23
Geometry Vol I to IV by Hall and Stevens
Try with Hints
Let \(\angle APB= \theta\)
area of \(\Delta PAB\)=\(\frac{1}{2}(3)(4)sin\theta\)
area of \(\Delta PAD\)=\(\frac{1}{2}(3)(6)sin(\pi-\theta)\)
area of \(\Delta PDC\)=\(\frac{1}{2}(8)(6)sin{\theta}\)
area of \(\Delta PCD\)=\(\frac{1}{2}(8)(4)sin(\pi-\theta)\)
or, area of quadrilateral ABCD is \(\frac{1}{2}(12+18+48+32)sin{\theta}\)
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.
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
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........
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
