2012 MOCK AIME released! (Geometry)

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1080 posts

#1 h Jun 17, 2012, 7:41 PM • 20 Y

Y by Diehard, trigonometry456103, Cyclicduck, airplanes1, fermat007, mymathboy, AkshajK, tc1729, Binomial-theorem, dantx5, mathgenius64, AwesomeToad, Amir Hossein, teethpaste, parmenides51, Adventure10, Mango247, and 3 other users

All right! After a few long weeks of compiling, revising, deleting, compiling, deleting again…We finally have finished the MOCK AIME! Note that this is definitely one of the hardest AIME’s ever written, partly because of the sheer amount of geometry necessary and it is comprised of tedious work. Do not be disheartened if you cannot manage to finish all of them; I respect those who can even do one problem; it is an accomplishment. Also because of this, the USAMO process may be a bit flawed, and therefore, if the results of the mock AIME do not reflect on true results, a.k.a. being too hard, we will create a shorter, much easier, proof oriented competition, similar in many ways to USAMO.

You will have 2 weeks to complete this competition. If you are struggling to meet this deadline at the fact that you are on vacation, have math camps, or something else like that, please PM me, and we will discuss making your submission deadline longer. The deadline is from now- 6/17/12-7/1/12- July 1st, 2012. Any submissions after this date will be void, unless you have PMed me.

We are hoping that you will treat this like the real AIME, and spend 3 hours on it, but we have no way in enforcing that. Please just remember that you are competing against other AOPSers, and honor is important.
When you are finished, please PM your answers in the format below.
Ex. Subject: MOCK AIME ANSWERS USER: (USER)
Body:
1. 244(ANS)
2. 244(ANS)
3. …
4. …

All the way to 15. Note that 244 is NOT the answer, just a test charge. We recommend that even if you do not know how to do the problem, please submit an answer for every single one, even if you are just guessing. That way, you still have a 0.01% chance at getting it right. PM them to me!

Please additionally note that since this is a mock AIME, the answers will be in the format the positive integers between 1-999, inclusive. So if your answer turns out to be something different, please PM me if you think there is a mistake.
We wish you the best of luck, and hope you do your best!

This thread will be locked until the end of the testing period. Please PM me if you have any questions or concerns. I will be posting the solutions at the end of the testing period.
EDIT: This thread will NOT be locked.

A huge thanks to Diehard and 1=2 for everything. This could never have happened with them two. Even a greater thanks to Diehard for creating everything in the first place! Thanks to 1=2 especially because of (his) amazing problem skills.

Problem 1. The area of the circle can be expressed as $\frac{\pi}{n}$ , for some positive integer $n$ . Compute $n$ .
Change to number 4: Find $EF^2$ .
Number 5 is in degrees
Problem 6. Instead of "Find the ratio of the areas of these two triangles", it should say "The ratio of the areas of these two triangles can be expressed as $m+n\sqrt{p}$ , where $m$ and $n$ are positive integers and $p$ is a prime. Compute $m+n+p$ .
Problem 10. The area of region $R$ can be expressed as $n\pi$ , for some integer $n$ . Compute $n$ .
#12: Take the answer mod 1000
Number 15 is integral.

Apologies for any inconviences this has caused.

I have attached the NEW AIME. This is the edited version, and I have changed all the mistakes. If you STILL think there are mistakes, PM me immediately.

Also note that I will accept new answers if you have PMed me the answers already, due to the mistakes.

Attachments:

2012 MOCK AIME.pdf (33kb)

EDITED AIME.pdf (35kb)

This post has been edited 10 times. Last edited by ProblemSolver1026, Jul 4, 2012, 2:45 PM

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Diehard

1374 posts

Diehard

#2 h Jun 17, 2012, 7:46 PM • 4 Y

Y by Cyclicduck, fermat007, Adventure10, Mango247

Good luck everyone.

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trigonometry456103

349 posts

trigonometry456103

#3 h Jun 17, 2012, 7:47 PM • 2 Y

Y by Adventure10, Mango247

thanks for doing this! I will do it immediately after I come back

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exmath89

2572 posts

exmath89

#4 h Jun 17, 2012, 8:49 PM • 2 Y

Y by Adventure10, Mango247

I think AIME answers range from 0 to 999. But thanks for creating this!

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ProblemSolver1026

1080 posts

ProblemSolver1026

#5 h Jun 17, 2012, 10:22 PM • 5 Y

Y by Seedleaf, Cyclicduck, tim9099xxzz, Adventure10, and 1 other user

EDIT: Change to question number one. Add this onto the end: For $\frac{\pi}{n}$ , find n.

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newchie123

412 posts

newchie123

#6 h Jun 17, 2012, 11:19 PM • 2 Y

Y by Adventure10, Mango247

very nice problems, they are quite tough for me, but really nice nevertheless

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ProblemSolver1026

1080 posts

ProblemSolver1026

#7 h Jun 17, 2012, 11:23 PM • 3 Y

Y by Diehard, Adventure10, Mango247

I would recommend everyone who signed up to have a try and go for it; even if you think they're hard. It's good experience.

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AkshajK

4820 posts

AkshajK

#8 h Jun 17, 2012, 11:36 PM • 1 Y

Y by Adventure10

Lol.

1. Geometry is my worst math subject
2. This AIME is one of the hardest AIMEs
3. This AIME requires lots of tedious work

(1) + (2) + (3) =

[i haven't looked at the test yet]

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ProblemSolver1026

1080 posts

ProblemSolver1026

#9 h Jun 18, 2012, 12:05 AM • 2 Y

Y by Diehard, Adventure10

. Geometry is definitely the hardest subject.

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Diehard

1374 posts

Diehard

#10 h Jun 18, 2012, 12:14 AM • 2 Y

Y by AwesomeToad, Adventure10

No combo

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ProblemSolver1026

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ProblemSolver1026

#11 h Jun 18, 2012, 12:19 AM • 2 Y

Y by Diehard, Adventure10

Additional notes:

Number 15 is integral.
Change to number 4: Find $EF^2$ .
Number 5 is in degrees.

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Diehard

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Diehard

#12 h Jun 18, 2012, 12:53 AM • 1 Y

Y by Adventure10

Number 15 is integral.
Change to number 4: Find $EF^2$ .
Number 5 is in degrees
#12: take the answer mod 1000

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1=2

4594 posts

1=2

#13 h Jun 18, 2012, 12:54 AM • 3 Y

Y by Seedleaf, NewAlbionAcademy, Adventure10

Problem 1. The area of the circle can be expressed as $\frac{\pi}{n}$ , for some positive integer $n$ . Compute $n$ .
Change to number 4: Find $EF^2$ .
Number 5 is in degrees
Problem 6. Instead of "Find the ratio of the areas of these two triangles", it should say "The ratio of the areas of these two triangles can be expressed as $m+n\sqrt{p}$ , where $m$ and $n$ are positive integers and $p$ is a prime. Compute $m+n+p$ .
Problem 10. The area of region $R$ can be expressed as $n\pi$ , for some integer $n$ . Compute $n$ .
#12: Take the answer mod 1000
Number 15 is integral.

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MCL

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MCL

#14 h Jun 18, 2012, 3:19 PM • 1 Y

Y by Adventure10

You guys are in need of editors. I can help out with that next time if you would like.

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trigonometry456103

349 posts

trigonometry456103

#15 h Jun 18, 2012, 3:57 PM • 2 Y

Y by Adventure10, Mango247

is there an error in #3? I keep getting a radical answer.

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roggin

133 posts

roggin

#187 h Jul 9, 2012, 12:20 AM • 2 Y

Y by Adventure10, Mango247

for problem 4 when you use ptlomy isnt it supposed to be $16a^{2} = 90$

since the diagnols are $4a$ and 6*12 + 6*3 = 90?

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numbertheorist17

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numbertheorist17

#188 h Jul 9, 2012, 9:02 PM • 3 Y

Y by girishvar12, Adventure10, Mango247

Can someone give solutions for number 15 and number 13? Thanks.

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Diehard

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Diehard

#189 h Jul 9, 2012, 9:32 PM • 1 Y

Y by Adventure10

This one is very very contrived... if someone has a different solution, please post it.

13. In $\triangle ABC$ , $29\leq{16cosA+24cosB+12cosC}$ . If $P,Q,R$ are extensions on sides $AB,BC,CA$ , respectively, of the triangle, and $|AP|=|BQ|=|CR|=1$ (these are vector quantities), then $xAP+yBQ+zCR=0$ (vector addition) where $x,y,z$ are positive integers and $(x,y,z)=1$ . Find $xyz$ .

Solution: We can prove the more general fact that for arbitrary real numbers $x,y,z$ and a $\triangle ABC$ , $x^{2}+y^{2}+z^{2}\geq{2zxcosA+2xycosB+2yzcosC}$ . This can be rewritten as $(xAP+yBQ+zCR)^{2}\geq{0}$ . Now, note that in this problem $x=4,y=3,z=2$ and the inequality is flipped. Hence $4AP+3BQ+2CR=0$ and $xyz=24$ .

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1=2

4594 posts

1=2

#190 h Jul 10, 2012, 2:39 PM • 2 Y

Y by Adventure10, Mango247

ProblemSolver1026 wrote:

Number 14:

Let $m\angle{B}=\alpha$ . Then the sum of the measures of the angles in the hexagon is : $720^\circ=10\alpha+5\alpha+5\alpha+2\alpha+\alpha+\alpha=24\alpha$

How do we know that $m\angle{E}=5m\angle{B}$ ?

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ProblemSolver1026

1080 posts

ProblemSolver1026

#191 h Jul 10, 2012, 3:34 PM • 1 Y

Y by Adventure10

1=2 wrote:

ProblemSolver1026 wrote:

Number 14:

Let $m\angle{B}=\alpha$ . Then the sum of the measures of the angles in the hexagon is : $720^\circ=10\alpha+5\alpha+5\alpha+2\alpha+\alpha+\alpha=24\alpha$

How do we know that $m\angle{E}=5m\angle{B}$ ?

We have $2m\angle{E}=10m\angle{B}$ , therefore we have $m\angle{E}=5m\angle{B}$ .

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nivle

114 posts

nivle

#192 h Jul 11, 2012, 4:48 PM • 2 Y

Y by Adventure10, Mango247

In the question, you only state: $m\angle{A} = 2m\angle{C} = 5m\angle{D} = 10m\angle{B} = 10m\angle{F}$
No angle E is mentioned in this problem.
I just let it equal $5m\angle{B}$ to make each angle a multiple of 30, which worked.

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numbertheorist17

268 posts

numbertheorist17

#193 h Jul 11, 2012, 6:13 PM • 3 Y

Y by girishvar12, Adventure10, Mango247

Using isosceles triangles and congruent triangles, you can prove that $\angle E=\angle C$ .

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ProblemSolver1026

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ProblemSolver1026

#194 h Jul 13, 2012, 2:09 AM • 1 Y

Y by Adventure10

numbertheorist17 wrote:

Using isosceles triangles and congruent triangles, you can prove that $\angle E=\angle C$ .

Indeed, you can do it this way. It's purely a coincidence that I forgot to put the $m\angle{E}$ part; but you can still find it out.

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numbertheorist17

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numbertheorist17

#195 h Jul 16, 2012, 5:21 PM • 3 Y

Y by girishvar12, Adventure10, Mango247

Could someone please post a solution for #15? Thanks.

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kevinnoh

206 posts

kevinnoh

#196 h Jul 18, 2012, 6:24 AM • 1 Y

Y by Adventure10

im confused... for #14 i got the answer for the area of hexagon ABCDEF Click to reveal hidden text but when i summed m and n i got 1/2 of the answer of the posted one...

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nivle

114 posts

nivle

#197 h Jul 18, 2012, 4:53 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

There was an edit for the question, so you had to find 2(m+n)
That sucks.

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mymathboy

163 posts

mymathboy

#198 h Oct 18, 2012, 4:47 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

ProblemSolver1026 wrote:

Heh. I should post the official answers now:

1.
2.
...
15. 12

Invalid image file

Invalid image file

For problem #15:

The solution $x = 12$ does not satisfy the condition that $AB = DC$ , since

$\frac{AB}{DC}=\frac{AB}{BD}\frac{BD}{DC}$

$=\frac{\sin{\angle{ADB}}}{\sin\angle{A}}\frac{\sin{\angle{C}}}{\sin\angle{DBC}}$

$=\frac{\sin{96^o}}{\sin{48^o}}\frac{\sin{36^o}}{\sin{60^o}} \~= 0.908299$

Numerically, when $x=12.9346^o$ , AB and DC are about to be equal.

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r00tsOfUnity

695 posts

r00tsOfUnity

#199 h Nov 17, 2023, 11:00 PM • 1 Y

Y by alan9535

thoughts on this mock, is it a high quality? should i take it as prep for the official AIME exam?

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wikjay

237 posts

wikjay

#200 h Nov 17, 2023, 11:04 PM

Y by

this post went to the top and i thought it was a new mock lol.

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parmenides51

30652 posts

parmenides51

#201 h Nov 25, 2023, 5:11 PM

Y by

problems had been posted in separate threads in aops (HSM)

I have just created a related post collection here

enjoy / start solving

more geometry mocks here,
more AIME mocks here,
more user-created contests here (contains all above)

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