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11/4 Meeting 1: Splitting Circles Equally by Area and Perimeter Solution: 2: Logic Questions about Knaves and Jokers Solution: 3: AMC Question https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_18 Solution: 4: Solve for the Perimeter of 5 conjoined semicircles Solution: Solving for Diameter: ​https://www.youtube.com/watch?v=Xrx8oa8NeXI Once Diameter is Known, Perimeter can be calculated as such: 2(22) + 2(12) + 16 = 84→ side lengths 5(36π/2) = 5(18π) = 80π → semicircle arcs ∴ Perimeter is 84 + 80π 11/12 Meeting 1: Law of Sines Problem Solution: https://www.youtube.com/watch?v=J-X9IiaOxG8 Essentially: 1. a/sin(​∠​CBD) = e/sin(​∠​BDC) = d/sin(​∠​ACB) (using law of sines in triangles ABC and BCD) 2. ∠​ABD + ∠ ​ ​E = 180 degrees (opposite angles in inscribed quadrilateral (ABDE) sums to 180) 3. ∠​ABD = 180-​∠​E, ​∠​ACD = 180-​∠​E, therefore ​∠​ABD = ​∠​ACD (same opposite angles theorem) 4. can rewrite equation in step 1 to be a/sin(​∠​CBD) = d/sin(​∠​ACB) ​ ​CBD = ∠ 5. As ∠ ​ ​B - ∠ ​ ​ABD, ∠ ​ ​CBD = ∠ ​ ​B - (180-​∠​E) ⇒ ∠ ​ ​CBD = ∠​ ​B+​∠​E-180 ⇒ ∠ ​ ​CBD + 180 = ∠​ ​B+​∠​E 6. thus, sin(​∠​CBD + 180) = sin(​∠​B+​∠​E) = -sin(​∠​CBD) (as sin(x+pi) = -sin(x)) 7. As ∠​ ​ACB = ∠​ ​C- ∠​ ​ACD, ∠ ​ ​ACB = ∠ ​ ​C - (​∠​180-​∠​E), and ∠​ ​ACB + 180 = ∠​ ​C+​∠​E 8. thus, sin(​∠​ACB+180) = sin(​∠​C+​∠​E) = -sin(​∠​ACB) 9. equation in step 4 ⇒ a/-sin(​∠​B+​∠​E) = d/-sin(​∠​C+​∠​E), negatives cancel out to get answer 2: Colliding Ants Solution: 3: 3 angles in a star problem Solution: 1. Method 1: a. The sum of the interior angles of a star is equal to 180 degrees i. Proof can be found here: https://www.youtube.com/watch?v=K3wYAGNmWJI b. thus A + B + C + 50 + 30 = 180, then A + B + C = 100 4: Logic question about tiles Solution: Hints: 1. We know that Vesta needs to have 2 even tiles, limiting her to {2,4}, {2,6}, or {4,6} 2. We know that Ysabel’s 2 tiles have to contain at least 1 prime number, limiting her to {1,3}, {1,5}, or {1,7} 3. As Alain has the greatest amount of tiles he has 3, while the others have two 4. Alain’s consecutive condition requires him to go over at least one even number; as the number 2 is present in at least 2 of Vesta’s possible numbers, Alain’s range cannot include two, limiting it to {3,4,5}, {4,5,6}, and {5,6,7} 5. From the given ranges the correct solutions can be logiced out