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Cyclic Quattrocento (Fall 2021 AMC 12B #24)

January 23, 2022

Fall 2021 AMC 12B #24 Problem: Triangle

$ABC$ has side lengths

$AB = 11, BC=24$ , and

$CA = 20$ . The bisector of

$\angle{BAC}$ intersects

$\overline{BC}$ in point

$D$ , and intersects the circumcircle of

$\triangle{ABC}$ in point

$E \ne A$ . The circumcircle of

$\triangle{BED}$ intersects the line

$AB$ in points

$B$ and

$F \ne B$ . What is

$CF$ ? Solution: We can create a diagram to try to find unique relationships in this problem. By angle chasing in the cyclic quadrilateral

$ABEC$ given that

$AE$ is the angle bisector of

$\angle{BAC}$ we can find that

$\angle{BAE} = \angle{CAE} = \angle{BCE} = \angle{CBE}$ . This means that triangle

$BCE$ is isosceles and

$BE = CE$ , which may be useful later on. Now, we can focus on

$CF$ , which is the length we want to find. We can see that

$ACF$ has a cevian

$CB$ , so we can use Stewart's Theorem (we will "derive" it with the law of cosines here) to solve for CF in terms of other side lengths. We can let

$\angle{ABC} = \theta$ which means...

Flower Petals (Fall 2021 AMC 12B #15)

January 15, 2022

Fall 2021 AMC 12B #15 Problem: Three identical square sheets of paper each with side length

$6$ are stacked on top of each other. The middle sheet is rotated clockwise

$30^\circ$ about its center and the top sheet is rotated clockwise

$60^\circ$ about its center, resulting in the

$24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form

$a-b\sqrt{c}$ , where

$a$ ,

$b$ , and

$c$ are positive integers, and

$c$ is not divisible by the square of any prime. What is

$a+b+c$ ? Solution: We can see that the resulting shape is a "regular" 24-sided polygon with equal side lengths, and equal interior angles. So we can take a "petal" of the polygon that consists of 3 adjacent vertices, find the area, and multiply by 12. The key here is to find the area of the petal by subtracting areas from a corner of the original square paper. From this diagram of a petal inside a

$3 \times 3$ corner of a square, we can see that each petal has an area...

Don't Be Intimidated (Fall 2021 AMC 12B #5, 8, 13)

January 15, 2022

Fall 2021 AMC 12B #5 Problem: Call a fraction

$\frac{a}{b}$ , not necessarily in the simplest form, special if

$a$ and

$b$ are positive integers whose sum is

$15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions? Solution: We can list out all the special fractions:

$\{\frac{1}{14},\frac{2}{13}, \frac{3}{12}, \frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3}, \frac{13}{2}, \frac{14}{1}\}$ . Simplifying, we have the fractions

$\{\frac{1}{14},\frac{2}{13}, \frac{1}{4}, \frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},\frac{8}{7},\frac{3}{2},2,\frac{11}{4},4, \frac{13}{2}, 14\}$ . We see that we can only add fractions with the same denominator , so we only need to focus on

$\{\frac{1}{4},\frac{1}{2},\frac{2}{3},\frac{3}{2},2,\frac{11}{4},4, \frac{13}{2}, 14\}$ . From the whole numbers

$\{2,4,14\}$ , we can achieve

$4, 6, 16, 8, 18, 28$ . From the fractions with deno...

Complete Residue System modulo m (Fall 2021 AMC 12A #25)

December 29, 2021

Fall 2021 AMC 12A #25 Problem: Let

$m\geq 5$ be an odd integer, and let

$D(m)$ denote the number of quadruples

$(a_1, a_2, a_3, a_4)$ of distinct integers with

$1\leq a_i \leq m$ for all

$i$ such that

$m$ divides

$a_1+a_2+a_3+a_4$ . There is a polynomial

$q(x) = c_3x^3+c_2x^2+c_1x+c_0$ such that

$D(m) = q(m)$ for all odd integers

$m\geq 5$ . What is

$c_1$ ? Solution: First, we must find a way to account for the number of ordered quadruples

$(a_1, a_2, a_3, a_4)$ such that

$m$ divides

$a_1+a_2+a_3+a_4$ . We know that there are

$(m)(m-1)(m-2)(m-3)$ total ordered quadruples since all elements must be distinct. First we can list out the sums of

$m$ quadruples of the form

$(a_1+n, a_2+n, a_3+n, a_4+n)$ where

$n$ takes on the values from

$0$ to

$m-1$ and all elements of the quadruple are modulo

$m$ . This means that the

$m$ quadruples have the same difference between consecutive elements modulo

$m$ . If we let the sum

$S = a_1+a_2+a_3+a_4$ , we have $(a_1, a_2, a_3, a_4) \Rightarr...

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