7/25/2025 H 2013 12 A reflection notes

2013 AMC 12A Log

Incorrect Problems

• Problem 20 – Totally confused
• Problem 24 – Right idea but casework had too many cases
• Problem 25 – No idea how to start

Correct Problems

Problems solved correctly: 15, 16, 17, 18, 19, 21, 22, 23

• Problem 16 – Took too long
• Problem 23 – Angle chasing was hard
• Had to look at answer choices

7/18/25 H 2012 AMC 12 B reflection notes

17 wrong confused

20 wrong, time sink

21 wrong 

22 right of hands

23 wrong understood solution

24 wrong time sink casework

25 easy casework, right 

Similar to 2023 Mathcounts chapter sprint #30, but harder (level 2)

This question is similar to, but more difficult than the 2023 Mathcounts Chapter Sprint #30, which is as follows:

Consider the seven points on the circle shown. If George draws line segments connecting pairs of points so that each point is connected to exactly two other points, what is the probability that the resulting figure is a convex heptagon? Express your answer as a common fraction.

Try the question first before you read the solution down below.

Convex Heptagon Probability (7 points on a circle)

Label the points 1 – 7 clockwise around the circle. Each point must be joined to exactly two others, so the drawing is a 2-regular graph (a disjoint union of cycles).

1 . Count all edge–sets (2-regular graphs) on 7 labelled points

• One 7-cycle.
  Fix the cyclic order: the edges of a 7-cycle correspond to a permutation of the 6 points after 1, with direction ignored. Hence
  \( \dfrac{6!}{2}=360 \) distinct 7-cycles.
• One 3-cycle + one 4-cycle.
  
  1. Choose the 3-cycle: \( \binom{7}{3}=35 \).
  2. Orient the 3-cycle: \( (3-1)!/2 = 1 \) way.
  3. Orient the remaining 4-cycle: \( (4-1)!/2 = 3 \) ways.
  Total  \( 35 \times 1 \times 3 = 105 \) edge–sets.

Adding the two cases gives the total number of admissible drawings: \[ N_{\text{total}} = 360 + 105 = 465 . \]

2 . Count the favourable edge–set

Exactly one of those drawings is the perimeter \(1\!-\!2\!-\!3\!-\!4\!-\!5\!-\!6\!-\!7\!-\!1\), which yields a convex heptagon.

3 . Probability

\[ \boxed{\displaystyle \Pr(\text{convex heptagon})=\frac{1}{465}} \]

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Checks: the same counting method gives \(70\) total for 6 points (hexagon) and \(3507\) total for 8 points (octagon), agreeing with \(1/70\) and \(1/3507\) respectively.

7/12/2025 2011 AMC 12A H reflection notes

2012 AMC 12 A Reflection Notes

Correct Answers: Questions 15–21

• Questions: 15, 16, 17, 18, 19, 20, 21
• Note on #16: Took time on one question using the Law of Cosines

Unanswered but Attempted: Questions 22, 23

• Q22 & Q23:
• Able to solve with help from solution
• Need to improve:
  • Organized counting of objects
  • 3D visualization skills
• Q23 Additional Note:
• Not able to solve during the test
• Watched a video solution afterward—it made sense, but I wouldn’t have made that connection under time pressure

Unanswered: Questions 24, 25

• Q24:
• Not able to compare exponents correctly
• Had to look at the solution to understand
• It was a complicated question
• Q25:
• Unable to solve even after reviewing the solution
• Gaps in understanding—don’t think I could have arrived at that solution myself
• Struggled with 3D visualization and drawing the necessary graph

7/8/2025 2011 AMC 12B H reflection notes

Q16 → Wrong due to silly mistake

Skipped Q20

• Time-sink solution requires heavy observation and quick identification.
• Involves quick use of inscribed angles and arcs.

Q25 → Not enough time

Was not able to solve; did not understand substitution of
n → n-k in the solution + solution video.
Redo!

Q24 Notes:

• Easy algebra but looked difficult at first.
• Just required extra time to complete — able to solve afterwards.

Q15, Q17, Q18, Q19, Q21, Q22, Q23 → Correct

• Q17: Online solution took a long time
• I had the fastest solution : the solution was first finding g(f(x)) = 10x-1, then subbing in the 1 to get h_1(1)=9, then continuously subbing 9 back into 10x-1 so it becomes 9, 89, 899, 8999….
• Q19: Took a long time
• Found faster solution but took time to get to it: this was a least upper bound question for the slope which hits the next lattice point. If I had the answers, it would have been easier to substitute them back into and find which one would work, but I had hidden the answers. What I did next was find a few of the closest points to 102, 50 which the slope would first intersect and the slope would be as close to 1/2 as possible. After testing a few values in the form of (n/1)/(2n+1) and (n+1)/(2n) which give values close to 1/2, 50/99 was the least of these.