# Thread: AMC Math questions - Help?

1. ## AMC Math questions - Help?

There are three questions I'm having trouble solving and would appreciate if you guys could solve and explain why (some of these are from the Year 11 and 12 section of the AMC test).

1) Screenshot_2.jpg

2) Screenshot_3.jpg

Answer is between 7/16 and 1/2

3) Screenshot_1.jpg

2. ## Re: AMC Math questions - Help?

1) let r be radius of cone, R be radius of quarter circle
From quarter circle, L = r*theta, 2pi*r = R*pi/2, R = 4r
From cone, sintheta = r/R, hence sintheta = 1/4

3. ## Re: AMC Math questions - Help?

I'm gonna label the corners of the outer square with A,B,C and D to make it easier, with A being the top left corner. And the corners of the inner rectangle with E, F, G, H with E being the upper most corner. [My apologies, all these letters may seem confusing to memorise so I suggest writing them down on a diagram somewhere so the solution is a little easier to follow]

Let angle AED = theta. From the larger unshaded triangle, we can tell that tantheta = 2.
Since AED, HEF, and BEF are supplementary, they add to 180, and HEF equals 90 whilst AED is theta. So doing the algebra (I trust that you can do this part since this question is one where if you can see it, the maths is pretty simple), and using angle sum of triangle, EFB equals theta. Using the same proof (supplementary angles, angle sum of triangle) angle FGC is also theta. Using the tantheta identity in each of the unshaded white triangles, you can get tantheta = something. (For the upper triangle, 1/2 is already given to you as a side length.) Let the other adjacent side be x, and let
1/2 over x = 2. (From the tantheta=2 identity you proved earlier on) And solve to get 1/4. Now since its a square the side length must be 1, so 1-1/4 gives 3/4 as the opposite angle in tantheta=2 for the bottom one.

(Oh, just a quick reminder that tantheta = opposite/adjacent, so the fraction itself can equal 2. Don't put tan(fraction)=2!)

Repeat the process for the bottom one, and you should get 3/8. Then, using pythagoreas theorem in each of the triangles, you should get respectively, rt(5)/4 and 3rt(5)/8. (Remember to square root the squares! It's a common silly error :P)

Then multiply them. I got 15/32 which is between 7/16(7/16=14/32) and 1/2. (1/2=16/32)

4. ## Re: AMC Math questions - Help?

Originally Posted by _Anonymous
There are three questions I'm having trouble solving and would appreciate if you guys could solve and explain why (some of these are from the Year 11 and 12 section of the AMC test).

1) Screenshot_2.jpg

2) Screenshot_3.jpg

Answer is between 7/16 and 1/2

3) Screenshot_1.jpg

$\noindent \textbf{Q3.} Are you sure the answer is 25\pi - 25? I got 50 square centimetres.$

5. ## Re: AMC Math questions - Help?

Originally Posted by 1729
$\noindent \textbf{Q3.} Are you sure the answer is 25\pi - 25? I got 50 square centimetres.$
yeah I got 50 as well

6. ## Re: AMC Math questions - Help?

Originally Posted by 30june2016
yeah I got 50 as well
$\noindent Okay, so assuming 50 is correct, then this is how I did it: The upper left quadrant of the figure can be translated and rotated 90\degree anticlockwise onto the upper right part, and similarly the lower left part of the figure can be translated and rotated 90\degree clockwise onto the lower right part. \\\\ The figure simply becomes a rectangle with dimensions 5 by 10, with area 50 square centimetres. \\\\ (the four parts I made reference to can be seen by constructing two perpendicular lines connecting the opposite corners in the figure.)$

7. ## Re: AMC Math questions - Help?

Originally Posted by 1729
$\noindent Okay, so assuming 50 is correct, then this is how I did it: The upper left quadrant of the figure can be translated and rotated 90\degree anticlockwise onto the upper right part, and similarly the lower left part of the figure can be translated and rotated 90\degree clockwise onto the lower right part. \\\\ The figure simply becomes a rectangle with dimensions 5 by 10, with area 50 square centimetres.$
Ah, wow I did it differently and never thought of something as fast and simple as that.

But anyway, the much longer way I thought of for the question was:

I drew a 10x10cm square around the circle, and drew circles of 5cm radii as shown.

From there I just subtracted the green area ([square - circle]/2)

Then subtracted the 2 quarter circles (the navy) and got an answer of 50 HAHAHAHAHAHAHAHAH way more trek though

8. ## Re: AMC Math questions - Help?

25pi/4+25pi/4+(25-25pi/4)+(25-25pi/4)

9. ## Re: AMC Math questions - Help?

Thank you all for your help.

10. ## Re: AMC Math questions - Help?

Originally Posted by 1729
$\noindent Okay, so assuming 50 is correct, then this is how I did it: The upper left quadrant of the figure can be translated and rotated 90\degree anticlockwise onto the upper right part, and similarly the lower left part of the figure can be translated and rotated 90\degree clockwise onto the lower right part. \\\\ The figure simply becomes a rectangle with dimensions 5 by 10, with area 50 square centimetres. \\\\ (the four parts I made reference to can be seen by constructing two perpendicular lines connecting the opposite corners in the figure.)$
I sort of understand your method, but am having trouble visualising it. Is there any way you can draw it or show a diagram?

11. ## Re: AMC Math questions - Help?

Originally Posted by Mongose528

25pi/4+25pi/4+(25-25pi/4)+(25-25pi/4)
How did you get those numbers?

12. ## Re: AMC Math questions - Help?

The radius of the circle is 5, so the area is 25pi

Area of shape = Area of Quarter circle + Area of Quarter circle + (25 - Area of Quarter circle) + (25 - Area of Quarter circle)

25 is the area of a 5*5 square

The top left and bottom right form quarter circles

the top right and bottom left areas are given by 25 - Area of Quarter circle

13. ## Re: AMC Math questions - Help?

Originally Posted by 30june2016
yeah I got 50 as well
same, i was just afraid to say it cauz...lol

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