[/quote]
,
with 
,
lies strictly between 
and 
on side 
,
lies strictly between 
on side 
such that 
.
is 
,
and 
are relatively prime positive integers. Find 
.
,
.
with 
,
to the sides 
are 
,
is the orthocenter. 
is the midpoint of segment 
.
and 
intersect at 
.
from 
.
are colinear
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
.
prime and 
satisfying the equation: 
be positive real numbers with 
.
and 
satisfy 
,
is an even number.
,
,
for which the equality holds.
but ham not getting one thing that what's the motivation is it like previously determined which inequality we have to come up with then use gcd or we use gcd to go to that inequality .I searched for this on MSE also but got nothing
Y by centslordm
A roll of tape is 
inches in diameter and is wrapped around a ring that is 
inches in diameter. A cross section of the tape is shown in the figure below. The tape is 
inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest 
inches.
![[asy]
/* AMC8 P22 2024, revised by Teacher David */
size(120);
pair o = (0,0);
real r1 = 1;
real r2 = 2;
filldraw(circle(o, r2), mediumgray, linewidth(1pt));
filldraw(circle(o, r1), white, linewidth(1pt));
draw((-2,-2.6)--(-2,-2.4));
draw((2,-2.6)--(2,-2.4));
draw((-2,-2.5)--(2,-2.5), L=Label("4 in."));
draw((-1,0)--(1,0), L=Label("2 in.", align=(0,1)), arrow=Arrows());
draw((2,0)--(2,-1.3), linewidth(1pt));
[/asy]](//latex.artofproblemsolving.com/e/d/b/edbca8b1b4e34ef5fac9cb2ad93cbde53f05a803.png)
inches in diameter and is wrapped around a ring that is 
inches in diameter. A cross section of the tape is shown in the figure below. The tape is 
inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest 
inches.![[asy]
/* AMC8 P22 2024, revised by Teacher David */
size(120);
pair o = (0,0);
real r1 = 1;
real r2 = 2;
filldraw(circle(o, r2), mediumgray, linewidth(1pt));
filldraw(circle(o, r1), white, linewidth(1pt));
draw((-2,-2.6)--(-2,-2.4));
draw((2,-2.6)--(2,-2.4));
draw((-2,-2.5)--(2,-2.5), L=Label("4 in."));
draw((-1,0)--(1,0), L=Label("2 in.", align=(0,1)), arrow=Arrows());
draw((2,0)--(2,-1.3), linewidth(1pt));
[/asy]](http://latex.artofproblemsolving.com/e/d/b/edbca8b1b4e34ef5fac9cb2ad93cbde53f05a803.png)
This post has been edited 2 times. Last edited by pog, Oct 20, 2024, 8:33 PM
revolutions. So the average length of tape is 
Multiplying gives 
This is closest to 
,
![$\frac{\pi\left[{2^2} - {1^2}\right]}{dr} = \frac{3\pi}{0.015}= \frac{3 \dot 1000 \dot \pi}{15} = \frac{3 \dot 200 \dot \pi}{3} = 200\pi \approx 628 \boxed{B}$](http://latex.artofproblemsolving.com/c/f/5/cf54db1eee7a89aea2312919fbbc8ed6901cb3dc.png)
revolutions. The average length of tape is 
This is closest to B 
