AMC Questions (2 Viewers)

[kurt.physics]

Q1 answer is 160

Could you show your work

 

[conics2008]

Q1 answer is 160

lol this is what i got but the dodgy style lol... xD..... what is the right answer Kurt ?

 

[Iruka]

O.K, you made a little triangle that is right angled. it has hypotenuse 5 metres and one leg 4 metres, so we can figure out that the other leg is 3 metres. I assumed that the road is a constant 4 metre width (which seems reasonable to me).

...This question would be much easier to explain if some of the vertices were labelled with letters...

However, since the two sides of the road are parrallel to each other, the angle that the upper side of the road makes with the left hand side of the paddock is the same as the angle that the lower side of the road makes with the left hand side of the paddock. I figured out what the tan of this angle is from the 3-4-5 triangle and then apply it to the larger triangle (i.e, the lower corner of the paddock) to find the width.

 

[kurt.physics]

lol this is what i got but the dodgy style lol... xD..... what is the right answer Kurt ?

I dont have it, the paper didnt come with solutions, that why i posted it up on the forum hoping someone could give the answer

 

[kurt.physics]

O.K, you made a little triangle that is right angled. it has hypotenuse 5 metres and one leg 4 metres, so we can figure out that the other leg is 3 metres. I assumed that the road is a constant 4 metre width (which seems reasonable to me).

...This question would be much easier to explain if some of the vertices were labelled with letters...

However, since the two sides of the road are parrallel to each other, the angle that the upper side of the road makes with the left hand side of the paddock is the same as the angle that the lower side of the road makes with the left hand side of the paddock. I figured out what the tan of this angle is from the 3-4-5 triangle and then apply it to the larger triangle (i.e, the lower corner of the paddock) to find the width.

Did you deduce that the two were similar, if so, how?

 

[undalay]

Refer to diagram.
a b c are angles

b = 180 - (90 + a)
b = 90 - a

c = 180 - (90+b)
c = 180 - (180 - a)
c = a

thus big triangle similar to small triangle.
Using ratio of sides

We work out width is 32m

Pythagoras, hypotenus is 40m

40x 4 = 160

 

[Iruka]

Did you deduce that the two were similar, if so, how?

Both right angled triangles, with one other angle the same (see my reasoning above - properties of transversals falling on parallel lines) hence they are similar.

If you draw the little triangle on the left hand side of the picture, you might see what I mean more easily

 

[kurt.physics]

Refer to diagram.
a b c are angles

b = 180 - (90 + a)
b = 90 - a

c = 180 - (90+b)
c = 180 - (180 - a)
c = a

thus big triangle similar to small triangle.
Using ratio of sides

We work out width is 32m

Pythagoras, hypotenus is 40m

40x 4 = 160

Thanks alot!

 

[conics2008]

O.K, you made a little triangle that is right angled. it has hypotenuse 5 metres and one leg 4 metres, so we can figure out that the other leg is 3 metres. I assumed that the road is a constant 4 metre width (which seems reasonable to me).

...This question would be much easier to explain if some of the vertices were labelled with letters...

However, since the two sides of the road are parrallel to each other, the angle that the upper side of the road makes with the left hand side of the paddock is the same as the angle that the lower side of the road makes with the left hand side of the paddock. I figured out what the tan of this angle is from the 3-4-5 triangle and then apply it to the larger triangle (i.e, the lower corner of the paddock) to find the width.

yes you sound good to me, you got my vote =) God bless you =) well donee.... it would help with that little triangle.. the angel is tan 3/4 then 90 - tan^-1 3/4 = 3/4 .... soo the angel at the bottom is 3/4

there fore widith is = to 3/4 = 24 /x do the maths solve for x you get x = 32 well done =)

 

[Iruka]

Now, how about the other question?

 

[undalay]

Is it possible to do without finding angles?

 

[conics2008]

^^ how about a long break.. say about forever... i solved my rubics cube like 10 times while trying to solve this question lol..

 

[Iruka]

You don't need to find any angles other than what's given.

 

[undalay]

You don't need to find any angles other than what's given.

hint?

 

[undalay]

ohmyyy.... curved corners!

I'll also guess that the circle fragments add up to one full circle.
so +4pi
the sides still add up to 125


So A?
Am i close

edit: yer im pretty sure its 125 + 4pi

 

[Iruka]

OK, we can add up the straight part of the guards path (when he is walking parrallel to the sides of the quadrilateral) and we get 18+18+40+45. Now we only have to figure out what happens at the corners R, P and Q. If the guard is to stay exactly 2 metres away from the corners, he must walk in a circular arc of radius 2 metres. So all we have to do is figure out how many degrees of rotation he must swing through, altogether.

Now, if he started somewhere along the side PS and is walking clockwise around the quadrilateral, we should count turns in the clockwise direction as positive, and those in the anti clockwise direction as negative. In total, he must swing through 360 degrees to end up where he started facing in the same direction, however, at S he does an anticlockwise turn of 90 degrees - so figure it out from there - he must travel the equivalent of 450 degrees in the clockwise direction around the other three corners.

 

[undalay]

OK, we can add up the straight part of the guards path (when he is walking parrallel to the sides of the quadrilateral) and we get 18+18+40+45. Now we only have to figure out what happens at the corners R, P and Q. If the guard is to stay exactly 2 metres away from the corners, he must walk in a circular arc of radius 2 metres. So all we have to do is figure out how many degrees of rotation he must swing through, altogether.

Now, if he started somewhere along the side PS and is walking clockwise around the quadrilateral, we should count turns in the clockwise direction as positive, and those in the anti clockwise direction as negative. In total, he must swing through 360 degrees to end up where he started facing in the same direction, however, at S he does an anticlockwise turn of 90 degrees - so figure it out from there - he must travel the equivalent of 450 degrees in the clockwise direction around the other three corners.

ah yes i used my own diagram which didn't go in ahaha.
But yeah makes sense for the diagram kurt gave. thx iruka

edit: btw how do u do Q10 efficiently: http://www.amt.canberra.edu.au/wuamcqs04.pdf

 

[Iruka]

The second circle will be tangent to both the first circle and the right hand side of the rectangle (somewhere). Let r be the radius of the second circle

Drop perpendiculars from the centres of the two circles to the base of the rectangle. Let x be the distance between these two perpendiculars. (Clearly 2+x +r = 6, so x = 4-r).

Now if you draw a line through the centres of the two circles, it will also pass through the point of tangency. (So you will have a trapezium ).

One of the parallel sides of the trapezium has length 2, the other has length r. of the two non-parallel sides, one has length x and the other has length 2+r.

Now we can use pythagoras theorem to show that
x^2 = (r+2)^2 - (r-2)^2,

But we have another expression for x in terms of r which we can substitute for x to obtain a quadratic in r.

You can then solve this to get the answer.

 

[buchanan]

I got these from the year 9 & 10 Intermediate Division Competition Paper from the AMC 2002 - questions 24 and 27. I posted it up on the forum hoping someone could give the answer.

1. Farmer Taylor of Burra Creek is very annoyed. A 4m wide road has been put through one of his rectangular paddocks, splitting it into two. As a result, he has lost some of his land.

If all dimensions in the figure are in metres, how much land, in square metres, has he lost?

(A) 120 (B) 150 (C) 160 (D) 200 (E) 250

2. The walls of a castle form a quadrilateral PQRS as shown so that PQ = 40 m, QR = 45 m, RS = 20 m, SP = 20 m and ∠PSR = 90°.

A guard must walk outside the walls so that he is always 2 m from the nearest part of the walls. The guard started walking around the castle clockwise and finally arrived at the starting point. The length, in metres, of his walk was

(A) 125+4π (B) 121+5π (C) 125+5π (D) 121+6π (E) 125+6π

Spoiler

Official answers: 1. C 2. B

Official solutions attached.

 

[kurt.physics]

Spoiler

Official answers: 1. C 2. B

Official solutions attached.

Hey, where did you get these? could you please post up the answers to the last 10 questions in the 2002 AMC? Preferably with worked solutions?