John Zhang is a beast.

Wed Nov 25, 2009 12:39 pm, by math154

http://gregory_r__smith.totallyexplained.com/

The parts about Florida and 4th place in 9th grade are somewhat questionable though.

Also the one above him on the list is somewhat awkward.

USAMTS Year 21 Round 2

Wed Nov 25, 2009 10:00 am, by math154

Problem 4: I managed to overkill quite a bit. Razz
Problem 5: Eh, I submitted two solutions on pages 6-8 and then found a nicer solution on page 9 after submitting. Sad
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Unfortunately, I have nothing better to post about.

Sun Nov 22, 2009 6:18 pm, by math154

GPML Large School Team Test 10704 #10
Parallelogram ABCD is constructed such that AB\ge BC and AC\ge BD. If AC/BD = AB/BC, prove that AC/AB = BD/BC = \sqrt2.
Solution 1
Let p = AB\ge BC = q and m = AC\ge BD = n. Then we are given that
\frac {m}{n} = \frac {p}{q}\implies q = \frac {pn}{m}.
Recall that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals, so
m^2 + n^2 = 2p^2 + 2q^2 = 2p^2 + 2\left(\frac {pn}{m}\right)^2 \\ \\ \implies m^2(m^2 + n^2) = 2p^2(m^2 + n^2) \\ \\ \impli...

Solution 2
Draw the altitude CX to AB. Let CX = h and BX = x. Also let AB = m. Assuming without loss of generality that \angle ABC is not acute and that AB\ge BC, Pythagorean Theorem yields
\frac {\sqrt {h^2 + (m + x)^2}}{\sqrt {h^2 + (m - x)^2}} = \frac {m}{\sqrt {h^2 + x^2}} \\ \\ \implies m^2h^2 + m^2(m - x)^2...


Can anyone find a solution involving transformations of the square case? Or some more geometrical way to see the \sqrt2 invariant ratio?

Classic problems (mainly geometry)

Sat Nov 14, 2009 8:10 pm, by math154

[ Currently: for fun ]

If you care...
OK so at this contest the test writer(s) decided to put two classic problems in... some inheritance problem and the original 20-80-80 triangle problem. The first one was acceptable but not that great for a multiple choice test, but the second one definitely shouldn't have been there because obviously most people who got it right probably didn't know how to solve it. If the answer had been "none of the above" then maybe that would be better. The main problem is time though, because this problem is not meant to be speedsolved.

-------------------------------
Inheritance problem: Excellence in Mathematics Contest 2009.18
The wealthy mother died leaving an estate of G gold coins, where G is an integer greater than 30. She has D daughters and they inherit the coins as follows. The eldest receives 1 coin plus 1/k of those remaining, where k is an integer. The second eldest receives 2 coins plus 1/k of those remaining. The third eldest receives 3 coins plus 1/k of those remaining. This pattern continues for the rest of the daughters, except that the youngest receives all the coins that were not inherited by her sisters. If each of the daughters received the same number of coins, then D + G is
\textbf{(A)}6k\quad\textbf{(B)}k^2 + k\quad\textbf{(C)}k^2 + 1\quad\textbf{(D)}(k + 1)^2\quad\textbf{(E)}\text{None of these}
Natural interpretation, in my opinion
First, note that each daughter gets G/D coins. Here, we assume that the n^{th} eldest daughter first gets n coins and then 1/k of the remaining coins (not including the n coins). Because n - 1 daughters each get G/D coins before the n^{th} daughter goes, we find that for 1\le n\le D - 1,
\frac {G}{D} = n + \frac {G - (n - 1)\frac {G}{D} - n}k \\ \\ \implies G((k - 1) - D) = n(D(k - 1) - G).
The wording implies that there are at least two daughters. If there are only two daughters, then D = 2 and the only valid value of n is 1, so
G(k - 3) = 2(k - 1) - G\implies(G - 2)(k - 2) = 2,
but clearly then G\le30.

However, if there are at least three daughters, then if D(k - 1) - G is nonzero, then n can only take on one value, while it must take on at least n - 1\ge3 - 1 = 2, which is impossible. Hence
D(k - 1) - G = G((k - 1) - D) = 0,
so because G > 30, k - 1 = D and G = D(k - 1) = (k - 1)^2, and hence
D + G = (k - 1) + (k - 1)^2 = k^2 - k\implies\boxed{\textbf{(E)}\text{None of these}}.

Other interpretation
Again, each daughter gets G/D coins. Here, we assume that the n^{th} eldest daughter first gets 1/k of the remaining coins before receiving n more coins. Because n - 1 daughters each get G/D coins before the n^{th} daughter goes, we find that for 1\le n\le D - 1,
\frac {G}{D} = n + \frac {G - (n - 1)\frac {G}{D}}k \\ \\ \implies G((k - 1) - D) = i(Dk - G) = 0
The wording implies that there are at least two daughters. If there are only two daughters, then D = 2 and the only valid value of n is 1, so
G(k - 3) = 2k - G\implies(G - 2)(k - 2) = 4,
but clearly then G\le30.

However, if there are at least three daughters, then if Dk - G is nonzero, then n can only take on one value, while it must take on at least n - 1\ge3 - 1 = 2, which is impossible. Hence
Dk - G = G((k - 1) - D) = 0,
so because G > 30, k - 1 = D and G = Dk = k^2 - k, and hence
D + G = (k - 1) + (k^2 - k) = k^2 - 1\implies\boxed{\textbf{(E)}\text{None of these}}.

-------------------------------
20-80-80 triangle: Excellence in Mathematics Contest 2009.20
Triangle ABC is isosceles (AB = AC). Point D is on side AC, such that \angle DBC = 60^\circ. Point E is on side AB, such that \angle ECB = 50^\circ. The triangle is shown in the figure but not to scale. If \angle BAC = 20^\circ then \angle AED, in degrees, is
\textbf{(A)}60\qquad\textbf{(B)}45\qquad\textbf{(C)}50\qquad\textbf{(D)}40\qquad\textbf{(E)}\text{None of these}
Solution
We prove below in problem 2 that \angle BDE = 30^\circ, so it's easy to see that
\angle AED = 50^\circ\implies\boxed{\textbf{(C)}50}.

-------------------------------
Let \triangle ABC be isosceles with vertex A such that AB = AC, with point E on AB and point D on AC (unless stated otherwise).

Problem 1: Find \angle BDE if \angle DBC = 70^\circ and \angle ECB = 60^\circ.
Solution 1
Originally discussed here.

Equilateral triangles are awesome, so construct E' on AC such that \angle E'BC = 60^\circ (EE' is parallel to BC), and let the intersection of line E'B with EC be point F. Note that \triangle BFC is equilateral. Furthermore, F is equidistant from AB and AC by symmetry, so AF is precisely the angle bisector of \angle BAC. This means that \angle BAF = \angle FAE' = 10^\circ. Because AE' = E'B (since \angle BAE' = \angle ABE' = 20^\circ), \angle BE'A = \angle AE'B, and \angle DBE' = \angle FAE' = 10^\circ, we have that \triangle BDE'\cong\triangle AFE', so DE' = FE'. But because \triangle EE'F is equilateral, DE' = FE' = EE'. Recall that EE' is parallel to BC, so now
\angle EE'D = 80^\circ\implies\angle DEE' = \angle EDE' = 50^\circ,
and since
30^\circ = \angle BDE' = \angle EDE' - \angle BDE = 50^\circ - \angle BDE,
we finally arrive at \angle BDE = 20^\circ.

Solution 2
Found on the first post here, solution 1.

As in solution 1, construct E' on AC such that \angle E'BC = 60^\circ (EE' is parallel to BC), and let the intersection of line E'B with EC be point F. From problem 4 (solution 1), \angle BDA = 150^\circ\Longleftrightarrow AD = BC, and because we do have \angle ADB = 150^\circ, AD = BC. Then from equilateral \triangle BFC and this, we may let \alpha = AD = BC = CF = BF. Realize also that FE = EE' = E'F, AE = EC, and AE = AE', so
DE' = AE' - AD = AE' - \alpha = AE - \alpha = EC - \alpha \\ = EC - FC = EF = EE',
so DE' = EE' and because that EE' is parallel to BC,
\angle EE'D = 80^\circ\implies\angle DEE' = \angle EDE' = 50^\circ,
and since
30^\circ = \angle BDE' = \angle EDE' - \angle BDE = 50^\circ - \angle BDE,
we finally arrive at \angle BDE = 20^\circ.

-------------------------------
Problem 2: Find \angle BDE if \angle DBC = 60^\circ and \angle ECB = 50^\circ.
Solution 1
Equilateral triangles are good. Construct D' on AB such that \angle BCD' = 60^\circ and let the intersection of BD and CD' be O (\triangle OBC is equilateral). Note that \angle BCE = \angle BEC = 50^\circ so BC = BE, and because \triangle BOC is equilateral BC = BO. Draw EO; we see that \angle BEO = \angle BOE = 80^\circ because BO = BE, and \angle BOC = 60^\circ because \triangle OBC is equilateral. Because \angle D'OC = 180^\circ, \angle D'OE = 40^\circ. Basic angle chasing tells us that \triangle OED\cong\triangle D'ED\implies\angle BDE = \angle D'DE. But because D'D is parallel to BC, \angle ADD' = 80^\circ, and we also know that \angle BDC = 40^\circ, so
\angle BDD' = \frac12\angle BDD' = \frac12(180^\circ - 80^\circ - 40^\circ) = 30^\circ.

Solution 2
Construct F on AB so that \angle BCF = 70^\circ. First realize that by problem 1, \angle CFD = 20^\circ, so \angle FDA = 30^\circ. By problem 4, AF = BC. But because \triangle BCE is isosceles, BC = BE, so AF = BE. Note also that because \angle ABD = \angle BAD = 20^\circ, AD = BD so \triangle BDE\cong\triangle ADF and hence \angle BDE = \angle ADF = 30^\circ.

-------------------------------
Problem 3: Find \angle DBE if \angle DBC = 50^\circ and \angle ECB = 40^\circ.
Solution
This is pretty much a trivial setup compared to the others. Let F be the intersection of EC and DB. Because \angle DBC = \angle BDC = 50^\circ and \angle BFC = 180^\circ - \angle DBC - \angle ECB = 90^\circ so EC is perpendicular to BD, we conclude that BF = FD and because EF = EF, the Pythagorean Theorem gives us BE = ED, so \triangle BEF\cong\triangle DEF by SSS congruency and \angle BDE = \angle EDF = \angle EBF = 30^\circ.

-------------------------------
Problem 4: Find \angle ADC if D is on AB and AD = BC.
Solution 1
Previously posted here.

Reflect C across AB to get B', and then B across AB' to get C'. Essentially we have placed two triangles congruent to \triangle ABC next to it so that \triangle ACC' is equilateral (since it is isosceles with vertex angle 3\cdot20^\circ=60^\circ). With some simple angle chasing, we find that \angle CC'B'=20^\circ, \angle C'B'C=150^\circ, and \angle B'CC'=10^\circ. But because AC=C'C, AD=BC=C'B', and \angle CAD=\angle CC'B'=20^\circ, SAS congruence tells us that \triangle CAD\cong\triangle CC'B', and hence \angle ADC=\angle C'B'C=150^\circ.

Solution 2
Originally discussed here.

Recall solution 1 to problem 1 (just swap D and E, and E' and D' in that solution to make this easier to see). It's clear that AD=CF=FB=BC, so when \angle ABD=10^\circ\Longleftrightarrow\angle ADB=150^\circ, AD=BC. But if \angle ADB\ne10^\circ, then it's clear that AD will be shorter or greater than BC.

-------------------------------
Problem 5: Prove that CE = DF given that \angle DCB = 60^\circ, E is on AC, F is on AB, and C,E,F are all on a circle with center B.
Solution
Let the intersection of circle B and line DC be G, different from C. Note that BC = BE = BG = BF because they are all radii of circle B, so because BC = BE and \angle BCE = 80^\circ, and simple angle chasing (remember that \angle DCB = 60^\circ) yields \angle BEC = 80^\circ\implies\angle ECF = \angle EBC = 20^\circ. Because \angle ECG = 20^\circ is an inscribed angle, arc GE = 40^\circ and hence \angle GBE = 40^\circ\implies\angle DBF = 20^\circ. By more simple angle chasing, \angle GDF = \angle FGD = 40^\circ, so DF = FG. But because FG and EC subtend the same arc of 20^\circ, EC = FG = DF, as desired.

Construction for AIME 1985.14

Sat Oct 31, 2009 9:14 pm, by math154

[ Currently: Just because ]

AIME 1985.14
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned 1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Construction
Err this wording sucks but it's late. Looking for symmetry, we may let the lowest ten players each get 4.5 points from each other and also 4.5 points from the other fifteen players, who get 7 points from each other and also 7 points from the lowest ten players. Let the lowest ten players be a. If a_i beats a_{i + 1,\ldots,i + 4\bmod{10}} and ties a_{i + 5\bmod{10}} then each has 4.5 points. Let the other fifteen players be b. If b_i beats b_{i + 1,\ldots,i + 7\bmod{15}} then each has 7 points. Now, consider only relations between the two groups. Tie b_{i,\;1\le i\le5} with a_{2i - 1},a_{2i}, and let edges denote some b defeating some a. Then 6 = \deg b_{i,\;1\le i\le5} and 7 = \deg b_{i,\;6\le i\le15}. Also, \deg a = 15 - 1 - 4 = 10. Connect b_{i,\;1\le i\le5} to a_{j,\;j\ne2i\pm1,(2i + 1)\pm1\pmod{10}}, giving each 10 - 4 = 6 edges and 3 edges to each a thus far. Now for each of the remaining b_i,\;6\le i\le15, connect to a_{i + 1,\ldots,i + 7\mod{10}}, giving each 7 edges and 3 + 7 edges total for all a.

Summarizing, there is at least one possible construction:
    a_i beats a_{i + 1,\ldots,i + 4\bmod{10}} and ties a_{i + 5\bmod{10}}
    b_i beats b_{i + 1,\ldots,i + 7\bmod{15}}
    b_{i,\;1\le i\le5} beats a_{j,\;j\ne2i\pm1,(2i + 1)\pm1\pmod{10}}
    b_{i,\;6\le i\le15} beats a_{i + 1,\ldots,i + 7\mod{10}}
    all unmentioned relationships contain some a_i beating some b_j

USAMTS Year 21 Round 1

Thu Oct 15, 2009 2:15 pm, by math154

Problem 1: I got the right answer but my solution looks really cramped and ugly, and some parts might not make sense.
Problem 2: I might have a few small errors.
Problem 3: I think it's okay.
Problem 4: I didn't show that my constructions gave a convex hexagon.
Problem 5: I have typos in the inequality chain at the end.

I don't know how they grade so I can't really say much...

Edit: 5/5/5/4/5, as pretty much expected.
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Lol

Tue Sep 22, 2009 4:23 pm, by math154

[ Currently: not caring if others have posted this before ]

http://www.wolframalpha.com/input/?i=screw+you

And because I am bored, here is a problem from some GPML thing...

GPML Sprint Test 10904 #30
Find the maximum value of \frac {ab + 2bc + cd}{a^2 + b^2 + c^2 + d^2} when a,b,c,d\in\mathbb R are not all zero.
\textbf{(A)}\ 1\qquad\textbf{(B)}\frac32\qquad\textbf{(C)}\frac {\sqrt2 - 1}2\qquad\textbf{(D)}\sqrt2\qquad\textbf{(E)}\text{...

Hint
There is a fun elimination solution. Darn, giving answers is not good.

Hints for a Legit Solution
Cauchy, Trig, or Calculus

Sigh...

Fri Sep 18, 2009 4:36 pm, by math154

I can't find the motivation to write a short story about anything.

Edit: Thanks... Now I will just put the link here for future reference because I don't want to bookmark this.

http://www.seventhsanctum.com/

Orchestra practice records...

Sun Sep 06, 2009 8:40 am, by math154

Now it will be a random integer picking assignment.

Note to self: Remember to change the stuff each week accordingly. It is somewhat annoying that they are actually using pdf's now...

Edit: In the future I will try to remember that it's difficult to practice on a day before the day or a viable practice time on a day is available...

Also, for all the copying, it is convenient to paste all the random.org stuff onto notepad separated by days for the second section and then copy it onto word.

Part 1: With this, pick the first two days (to have lots of homework) from the random ordering of this list.
Monday
Tuesday
Wednesday
Thursday

Part 2: Write down 60 minutes for Friday, Saturday, and Sunday, and 30 for the other two days (or whatever).
Part 3: With this, write these in the order they appear for the five days from the random ordering of this list.
Today I practiced scales, arpeggios,
I worked on scales, arpeggios,
During this practice session I focused on scales, arpeggios,
Today I did scales, arpeggios,
I mainly practiced scales, arpeggios,
I refined my scales, arpeggios,
Today I got better at scales, arpeggios,


For each day of practice:
Part 4: With this, write down the first three things from the random ordering of this list.
Suzuki 4
Suzuki 5
Suzuki 6
Dvorak
Bizet
Sibelius

(Adjust as necessary...)
Part 5: After this, write down "I improved..." or something else.
Part 6: With this, write down the first two things from the random ordering of this list.
intonation
bowing
tone quality
articulation
rhythm
fingering
style
dynamics
contrast

Part 7: Then write "I am having trouble with..." or something else.
Part 8: Now write the next two things on the list from Part 6.

It is probably easier to bs, but sometimes it is hard to be creative in which case this helps (it turns a creative writing assignment into a copy and paste assignment. Yay!)
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Whattttttttttttttt

Wed Jul 29, 2009 11:00 am, by math154

TIP was pretty weird and fairly fail...Duke East got shut down and I got to leave Davidson early for similar reasons (but it wasn't shut down). Yay. The only cool part was the fact that some people were beastly. There was a NINJA KID who was obsessed with ninjas and dressed up as a ninja sometimes (pretty legit). I'm too lazy to list other people.

Also for anyone who cares, this year the password to Math Problem Solving FTW games was mprob2009. I suspect it will be mprob2010 next year. Shocked I know this because I was in the computer lab while they were playing FTW.

DC was very awesome, though.

Sorry for removing contributors...ask if you want to be contributed.

My new favorite problem

Mon Jun 29, 2009 5:30 pm, by math154

USAMO 1984.5

P(x) is a polynomial of degree 3n such that

P(0) = P(3) = \cdots = P(3n) = 2
P(1) = P(4) = \cdots = P(3n - 2) = 1
P(2) = P(5) = \cdots = P(3n - 1) = 0
P(3n + 1) = 730.

Find n.

Solution
Since we all know that brute force is the way to go, let's use finite differences! Very Happy

\Delta^0 = \{\ldots,1,0,2,1,0,2,730\}
\Delta^1 = \{\ldots, - 1, - 1,2, - 1, - 1,2,728\}
\Delta^2 = \{\ldots, - 3,0,3, - 3,0,3,726\}
\Delta^3 = \{\ldots, - 6,3,3, - 6,3,3,723\}
\Delta^4 = \{\ldots, - 9,9,0, - 9,9,0,720\}
\Delta^5 = \{\ldots, - 9,18, - 9, - 9,18, - 9,720\}
\Delta^6 = \{\ldots,0,27, - 27,0,27, - 27,729\}
\Delta^7 = \{\ldots,27,27, - 54,27,27, - 54,756\}
\Delta^8 = \{\ldots,81,0, - 81,81,0, - 81,810\}
\Delta^9 = \{\ldots,62, - 81, - 81,62, - 81, - 81,891\}
\Delta^{10} = \{\ldots,243, - 243,0,243, - 243,0,972\}
\Delta^{11} = \{\ldots,243, - 486,243,243, - 486,243,972\}
\Delta^{12} = \{\ldots,0, - 729,729,0, - 729,729,729\}

And the dudes are like equal so we are done! 3n = 12\implies n = \boxed4 (the degree can't be higher since that would require the stuff to be eventually constant but since 730\ne1, necessary for it to be constant since it's a periodic cycle, it's impossible). The unfortunate thing is that this is pretty darn fast.

Yay

Sat Jun 06, 2009 1:23 pm, by math154

Just got PSS, ACoPS, Geo Revisited, Math Oly Treasures and Challenges. Also just finished AoPS 2, so it was good timing.

Finally

Thu May 28, 2009 1:14 pm, by math154

Outta middle school!

Science Olympiad!

Tue May 12, 2009 6:07 pm, by math154

Yeah....good luck if anyone is participating.

MATHCOUNTS (yeah)

Mon May 11, 2009 12:54 pm, by math154

So uh...highlights of MATHCOUNTS?

Thursday: Came in to get a nice greeting from LA. Also got owned by levans. Razz Ironically, this may have motivated me. Shocked Max too.

Friday: Nothing special here. The Magic Kingdom was good I guess? Played fail poker with random people (including Max) in Runpeng's and my room.

Saturday: Epcot was okay as well. The banquet was interesting. Missouri was like yay epic win when we were assigned with this one team. I got to sit next to David. Pwnage. Then everyone went in to our room again, and we played fail mafia. For like 0 seconds. Then it got pretty disturbing after Max left. Kathleen and Matt were sorta acting weird. Runpeng fell asleep after we moved to like 2 other rooms so Siddhant got to carry him back to our room. Went to sleep at like 10:30...the next morning in the plane.

That's all.

Clarification: I had an all-nighter? I did say the next morning anyway.

Math class

Wed May 06, 2009 12:12 pm, by math154

In my Algebra II test today, I had to "find five angles with 'reference angle' 54^\circ, of which at least two are negative." I didn't really know what "reference angle" meant from a lack of paying attention, so I decided to guess that it just meant 54^\circ\pm360k^\circ. So I subtracted 360^\circ two times and got - 666^\circ! Then I realized 360|666000 so my answers were - 666^\circ, - 666666^\circ, - 666666666^\circ, - 666666666666^\circ, - 666666666666666^\circ.

Edit: Yes apparently I got full credit for this problem.

USAMO tomorrow!

Mon Apr 27, 2009 7:58 pm, by math154

Good luck to everyone! Smile

Stupid algebra

Fri Apr 24, 2009 3:56 pm, by math154

Factor a^4 - b^4 + c^4 - d^4 - 2(a^2c^2 - b^2d^2) + 4ac(b^2 + d^2) - 4bd(a^2 + c^2).

My Failing Solution

a^4 - b^4 + c^4 - d^4 - 2(a^2c^2 - b^2d^2) + 4ac(b^2 + d^2) - 4bd(a^2 + c^2)

= (a^2 - c^2)^2 - (b^2 - d^2)^2 + 4ac((b - d)^2 + 2bd) - 4bd((a - c)^2 + 2ac)

= (a - c)^2(a + c)^2 - (b - d)^2(b + d)^2 + 4ac(b - d)^2 + 8abcd - 4bd(a - c)^2 - 8abcd

= (a - c)^2(a^2 + c^2 + 2ac - 4bd) - (b - d)^2(b^2 + d^2 + 2bd - 4ac)

= (a^2 + c^2 - 2ac)(a^2 + c^2 + 2ac - 4bc) - (b^2 + d^2 - 2bd)(b^2 + d^2 + 2bd - 4ac)

= (a^2 + c^2)^2 - 4bd(a^2 + c^2) - 4a^2c^2 + 8abcd - (b^2 + d^2)^2 + 4ac(b^2 + d^2) + 4b^2d^2 - 8abcd

= (a^2 + c^2) + 4ac(b^2 + d^2) + 4b^2d^2 - (b^2 + d^2)^2 - 4bd(a^2 + c^2) - 4a^2c^2

= (a^2 + c^2 - 2bd)^2 - (b^2 + d^2 - 2ac)^2

= (a^2 - b^2 + c^2 - d^2 + 2ac - 2bd)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd)

= ((a + c)^2 - (b + d)^2)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd)

= \boxed{(a - b + c - d)(a + b + c + d)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd)}.

A geometry problem

Thu Apr 16, 2009 5:33 pm, by math154

In isosceles \triangle ABC with vertex angle m\angle A = 20^\circ, M is constructed on \overline{AB} such that AM = BC. Find m\angle AMC.

Source: learned from Kevin Li

Surprisingly Nice Solution

Look at the diagram, where we place two triangles congruent to \triangle ABC next to it so that \triangle ACC' is equilateral. m\angle CAM = 20^\circ by definition,

m\angle C'CB = m\angle ACB - m\angle ACC' = 80^\circ - 60^\circ = 20^\circ, and

m\angle BB'C' = 160^\circ so since BB' = B'C' = x, m\angle C'BB' = 10^\circ and

m\angle CBC' = m\angle CBB' - m\angle C'BB' = 160^\circ - 10^\circ = 150^\circ.

Because BC = AM = x, m\angle BCC' = m\angle MAC = 20^\circ, and

CC' = AC = x + y (CC' = x + y because \triangle ACC' is equilateral), we have

\triangle AMC\cong\triangle CBC' and

m\angle AMC = m\angle CBC' = \boxed{150^\circ}.


Maybe I should post my motivation here sometime: a two page trig solution using the triple angle identity as the crux move. Razz

Edit: Bash Yay

By the Law of Sines \frac {AM}{\sin(160^\circ - m\angle AMC)} = \frac {AC}{\sin(m\angle AMC)} and

\frac {BC}{\sin{20^\circ}} = \frac {AM}{\sin{20^\circ}} = \frac {AC}{\sin{80^\circ}}, so dividing gives

\frac {\sin{20^\circ}}{\sin(160^\circ - m\angle AMC)} = \frac {\sin{80^\circ}}{\sin(m\angle AMC)}. Let x = m\angle AMC so that

\sin x\sin{20^\circ} = \sin{80^\circ}\sin{160^\circ - x}

\implies2\sin{10^\circ}\cos{10^\circ}\sin x = \cos{10^\circ}\sin{x + 20^\circ}

\implies2\sin x\sin{10^\circ} = \sin x\cos{20^\circ} + \cos x\sin{20^\circ}

\implies\sin x(2\sin{10^\circ} - \cos{20^\circ}) = \cos x\sin{20^\circ}

\implies\sin^2x(4\sin^2{10^\circ} - 4\sin{10^\circ}\cos{20^\circ} + \cos^2{20^\circ})

= \cos^2x\sin^2{20^\circ} = \sin^2{20^\circ} - \sin^2{20^\circ}\sin^2x

\implies\sin^2x = \frac {\sin^2{20^\circ}}{\sin^2{20^\circ} + 4\sin^2{10^\circ} - 4\sin{10^\circ}\cos{20^\circ} + \cos^2{20^\...

= \frac {4\sin^2{10^\circ}\cos^2{10^\circ}}{4\sin^2{10^\circ}\cos^2{10^\circ} + 4\sin^2{10^\circ} - 4\sin{10^\circ}(1 - 2\sin...

= \frac {4\sin^2{10^\circ}(1 - \sin^2{10^\circ})}{4\sin^4{10^\circ} + 8\sin^3{10^\circ} + 4\sin^2{10^\circ}(1 - \sin^2{10^\ci...

= \frac {4\sin^2{10^\circ} - 4\sin^4{10^\circ}}{8\sin^3{10^\circ} + 4\sin^2{10^\circ} - 4\sin{10^\circ} + 1}.

Now we know that \sin{30^\circ} = \frac12 = 3\sin{10^\circ} - 4\sin^3{10^\circ}, so substitution yields the previous expression as

\frac {4\sin^2{10^\circ} - 4\sin^4{10^\circ}}{4\sin^2{10^\circ} + 2\sin{10^\circ}}

= \frac {2\sin^2{10^\circ}(1 - \sin^2{10^\circ})}{\sin{10^\circ}(1 + 2\sin{10^\circ})} = \frac {2\sin{10^\circ} - (2\sin^3{10...

= \frac {2\sin{10^\circ} - \left(\frac32\sin{10^\circ} - \frac14\right)}{1 + 2\sin{10^\circ}} = \frac14!

Thus \sin x = \pm\frac12, and since x = 30^\circ makes m\angle MCB = - 50^\circ by the Exterior Angle Theorem, we conclude that the only possibility is

x = \boxed{150^\circ}.


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Yay

Fri Apr 10, 2009 9:43 am, by math154

USAMO, 144/9=234 10A and 115.5/9=205.5 12B I think. Haven't gotten the official email yet.

Edit: Checking state results, the indices are correct.

AIME lI

Wed Apr 01, 2009 2:41 pm, by math154

I think I got a 9. [1,8] and [10]. I definitely should have had [1,11], but I did something with nonnegative/positive on number 9 that ended up giving me an answer of 998 instead of 1000. On number 11, I forgot that m\ge n, so I spent a long time on m < n cases and since (m,n) = (1,51) works, I got 176 instead of 125. This was probably harder than the AIME I, but I can't be sure. 144 on the AMC10A and a failing 115.5 on the AMC12B. So I made 2 careless mistakes on this, 1 on the 10A, and 6 or so on the 12B. Whatever...

Comments

1-4: Boring as usual.
5 was slightly better.
6 was really common; it was annoying but I managed to circumvent most arithmetic.
7 was okay but the arithmetic sucked.
8 was a nice problem I guess.
9 was okay as well.
10 was fine but pretty boring.
11 was fine except m\ge n was not large enough for me to see. Razz
12 was good taken out of the contest setting (finding the optimal configuration was cool).
13 was a nice problem.
14 was good I suppose, but calculus makes the insight very obvious...
15 was probably my favorite problem.


Time to worry about MATHCOUNTS again. Sad

Edit: I just realized, looking at my work, that I was going in a valid direction (not as nice as the trig substitution though), but actually blanked out on 14 and potentially may have gotten it, giving me a 12. But a 12 on the AIME I was much easier to get, so I'll just say the AIME II was harder.

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  • ok done

    By math154, on Mon Nov 23, 2009 8:11 pm

  • dude you should add me as contrib! Wink

    By ghjk, on Sun Nov 22, 2009 9:48 pm

  • hey wait what

    By Poincare, on Fri Oct 16, 2009 2:50 am

  • NOOOOOOOOOOOOOOOOOOOESSSS
    <end randomness>

    By abacadaea, on Tue Aug 11, 2009 5:54 pm

  • contribbbbb.

    By nikeballa96, on Wed Jul 29, 2009 12:46 pm

  • /me wants contrib

    By gauss1181, on Wed Jul 29, 2009 11:22 am

  • wasnt mewto like ur teammate or somethin? u should make him a contributor Mr. Green Mr. Green

    By draops, on Thu Jul 23, 2009 7:11 pm

  • um dude i should contrib????

    By Mewto55555, on Wed Jul 15, 2009 5:15 pm

  • uh that link just led to this blog Shocked

    By gauss1181, on Thu Jul 02, 2009 3:55 pm

  • yes indeed

    By math154, on Wed Jul 01, 2009 9:18 am

  • woah

    Algebra sucks

    By xpmath, on Tue Jun 30, 2009 6:12 am

  • hm 48=2^4\times3

    By gauss1181, on Sat Jun 06, 2009 3:12 pm

  • test

    This is a test

    By dnkywin, on Thu May 28, 2009 1:16 pm

  • actually 1*3*7*13*37

    By Anonymous, on Thu May 21, 2009 1:45 pm

  • lawl@hide tag

    anyway 10101=10^4+10^2+1=(10^2-10+1)(10^2+10+1)=91*111=(3*7)*(13*37) is pretty cool.

    By math154, on Wed May 20, 2009 5:29 pm

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