Track Math Quiz

To go further, I had to use trigonometry.

From x and/or y, you can use trigonometry to find the angle whose sin is x/r, or cosine is (36.5-y)/r, where r=36.5.

This angle, in radians, times the radius gives the answer:

0.65913*36.5 = 24.058

Bad Wigins wrote:

the brute force algebra

let s = 42.2, r = 36.5, j = 65. Look for x and y such that

x^2 + y^2 = j^2 and

(x-s)^2 + (y-r)^2 = r^2, x > s

2nd => (x^2 - 2sx + s^2 + y^2 - 2ry = 0

s^2 + j^2 = 2sx + 2ry

x = (s^2 + j^2 - 2ry)/2s

= 71.2 - .865y (kill it before it grows!)

1st => (71.2 - .865y - s)^2 + (y-r)^2 = r^2

(29.0 -.865y)^2 + y^2 - 73 = 0

...

y^2 - 70.5y + 481 = 0

solve for y and x, compute arctan and arc length etc etc.

many thanks for that

it was actually my fault for trying to attempt this sitting on the sofa, tv on & on back of an envelope

i sat down at at desk, turned off all noise & got out a proper piece of paper & did it manually with a substitution for y^2 as an x polynomial which landed a linear form for x & y which was easy to substitute back in the other equation & solving a standard quadratic

however, it was ugly & not recommended

rekrunner wrote:

Medium one first:

Area is 7648.612 m^2

4 laps around this area is 1466.4m

Rupp could run this in 3:27.72

Me? Harder to say. Maybe with proper training 4:45 to 5:25

What'd you do for this one, use the first derivative? I went down that path but got sidetracked and didn't have much time to spend.

I started with

short side of rectangle = x

long side of rectangle = 84.39+2y

#1 A(x)=x(84.39+2y)

#2 1/4x^2+y^2=36.5^2

i solved for y in #2, plugged into #1, found A'(x), but it got nasty from there. Probably went down the wrong path somewhere.

i was always told if you have something ugly, look for another method !!!

by far the most elegant way was by trig ( something i admit i was never very interested in, but it's growing on me ! )

main thing is to draw a decent straight line & curve ( did manage that on back of an envelope ! )

mark out the numbers given & draw some connecting lines to get some triangles

look at the triangles :

you see one where :

A = J + J'

where A is the angle of a large scalene with :

- opposite : 65

- adjacents of 36.5 & (36.5^2 + 42.195^2)^(1/2)

J ( for javelin !!! ) is angle we want & J' the compliment angle of abutting triangle with

- opposite 42.195

- adjacent 36.5

for A, use cosine rule on 3 known lengths ->

A = 86.905 degrees

J' = arctan ( 42.195 / 36.5 ) = 49.139 degrees

therefore

J = 37.765 degrees

& required length on curve is

(37.765/180) * ( 36.5 * pi ) =

24.058m

i'd have to say that is most elegant way of doing it

no x or y & no messy, miserable quadratics + it took about 2 minutes to solve once you'd drawn some nice triangles

What did I do? Wrote a program. But let's do it old school:I started with slightly different equations:1) A = (84.39 + 2y) (2x)2) x^2 + y^2 = 36.5^2Using 2), equation 1) can be rewritten solely in terms of y:3) A = (168.78 + 4y) (36.5^2 - y^2) ^ 1/2Now -- how to calculate dA/dy ? Could be messy, and it is, but we have rules:Product rule: d(u(y)*v(y)) = d(u)/dy * v(y) + d(v)/dy * u/dyChain rule: d(u(v(y))) = d(u)/dv * d(v)/dySwitching to prime notation to make the rest cleaner:First apply the product rule to 3):A' = [(168.78+4y)'] * (36.5-y^2)^(1/2) + [(36.5^2 - y^2)^(1/2)'] * (168.78 + 4*y) = (4) * (36.5-y^2)^(1/2) + [(36.5^2 - y^2)^(1/2)'] * (168.78 + 4*y)Then the chain rule on the remaining sqrt: = (4) * (36.5-y^2)^(1/2) + (1/2) * (36.5^2 - y^2)^(-1/2) * (-2y) * (168.78 + 4*y)Now it's just algebra to get: = (-8y^2 - 168.78*y +5329) / (36.5^2 - y^2)^(1/2)The interesting part is (-8y^2 - 168.78*y +5329), which can be solved by the quadratic formula:y = 17.3331627 (and also -38.4307 which we can reject for two reasons)then, plugging back into 2) :x = 32.12182235and then applying 1) :A = 7648.61227

hmkksd wrote:

What'd you do for this one, use the first derivative? I went down that path but got sidetracked and didn't have much time to spend.

I started with

short side of rectangle = x

long side of rectangle = 84.39+2y

#1 A(x)=x(84.39+2y)

#2 1/4x^2+y^2=36.5^2

i solved for y in #2, plugged into #1, found A'(x), but it got nasty from there. Probably went down the wrong path somewhere.

nice

I looked for a Euclidean cheap trick but I didn't see that.

euclid is never a cheap trick, as remembring from awhile

in my school days, they were top guys who brought it out, no matter how good at subs they were

it was a race

euclid faster -> win

kid never looks for euclid

but, i never did much than brute force, but as ole boy always look for

"elegance"

if that had been 1st problem in uni entry, you wouda spent ~ 5' - 10' on it bruting algebra

if you are still college kid, then euclid wouda done it in 3' - 5', saving you :

helluva lotta minutes for next problem !

ventolin^3 wrote:

euclid faster -> win

only if a simple geometric solution exists.

this was always "simplest/hardest" problem i ever saw :

"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

solve it & i'll buy you serious single-malt at confluence...

Bad Wigins wrote:only if a simple geometric solution exists

your brain has to look for it simultaneously

you are a good "brute" mathematician

but i was probably close 35+y years ago, but coudn't solve 'em fast/ELEGANT enough back then

i'm an ole boy, so no fight here :

the aim of problem was to solve it quick in exam conditions in order to proceed :

you failed...

ventolin^3 wrote:

"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

solve it & i'll buy you serious single-malt at confluence...

16.21 feet.

HardLoper wrote:Random thought... 84.39 is exactly 1/500th of a marathon

After all that hard training the big race arrived - the World Championship Short-track Marathon. 250 out-and-back laps of the 84.39m homestretch.

100 runners lined up at the start. Amazingly none of them collided and all of them finished. Many of them were lapped, 168.78m or more behind. Despite the short laps no lapped runner came back to win.

All of them ran at a perfectly even pace, and no two of them ever reached one end of the course simultaneously.

How many times did two runners pass each other during the race, either in the same direction or in different directions? Count all passes for all possible pairs of runners.

(hmkksd was me, forgot to change my handle back)

Well I got the derivative calculated correctly, which was the hard part I guess. I'm proud for just getting that far, been years since I've done any math other than basic arithmetic.

Is this the nerdiest thread on letsrun ever?

Bad Wigins wrote:

After all that hard training the big race arrived - the World Championship Short-track Marathon. 250 out-and-back laps of the 84.39m homestretch.

100 runners lined up at the start. Amazingly none of them collided and all of them finished. Many of them were lapped, 168.78m or more behind. Despite the short laps no lapped runner came back to win.

All of them ran at a perfectly even pace, and no two of them ever reached one end of the course simultaneously.

How many times did two runners pass each other during the race, either in the same direction or in different directions? Count all passes for all possible pairs of runners.

Just a quick answer here, no time to see if it checked out.

Each runner passes all 99 others on a typical straightaway.

The first runner passes no-one on the first straight. The last runner passes no-one on the last straight. Similarly, A runner in the middle doesn't pass the runners behind on the first straight, or the runners ahead on the last. So each will pass all the others 499 out of 500 straights (whether the 1st 499 or last 499)

A lapped runner will add 2 for that straight, but subtracts 1 for the last two straights since the other runner finishes a full lap ahead. Keeping the average at 99.

499straights * 99passes/straight/runner = 49,401 passes/runner.... (*100 runners)

each pass of two runners only counts once. You're counting many of them twice, one for each runner.

Once you figure this out, just post the answer, so others can try to solve it too.

I am thinking 2,470,050. Good thread, I hope people keep it alive.

That is deceptively hard. The math gets messy quickly.

ventolin^3 wrote:

this was always "simplest/hardest" problem i ever saw :

"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

solve it & i'll buy you serious single-malt at confluence...

I counted 2475000.Note I counted the start as the first pass.

Bad Wigins wrote:

After all that hard training the big race arrived - the World Championship Short-track Marathon. 250 out-and-back laps of the 84.39m homestretch.

100 runners lined up at the start. Amazingly none of them collided and all of them finished. Many of them were lapped, 168.78m or more behind. Despite the short laps no lapped runner came back to win.

All of them ran at a perfectly even pace, and no two of them ever reached one end of the course simultaneously.

How many times did two runners pass each other during the race, either in the same direction or in different directions? Count all passes for all possible pairs of runners.

What would the total be if one runner was lapped once by all the others, but then sped up (still to an even pace) and came back to win?

rekrunner wrote:That is deceptively hard. The math gets messy quickly

ventolin^3 wrote:

this was always "simplest/hardest" problem i ever saw :

"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

solve it & i'll buy you serious single-malt at confluence...

i did say so

fortunately, i have seen a solution for all the mess :