23S=131−13two-thirds cap S equals the fraction with numerator one-third and denominator 1 minus one-third end-fraction
BA2=BD×BKcap B cap A squared equals cap B cap D cross cap B cap K is the second intersection point on line BCcap B cap C
S−13S=13+(29−19)+(327−227)+(481−381)+…cap S minus one-third cap S equals one-third plus open paren two-nineths minus one-nineth close paren plus open paren 3 over 27 end-fraction minus 2 over 27 end-fraction close paren plus open paren 4 over 81 end-fraction minus 3 over 81 end-fraction close paren plus …
Scratch paper and a writing utensil only. Calculators are strictly prohibited. Mathcounts National Sprint Round Problems And Solutions
Now multiply by C-counts: S=9: 9 pairs × 2 C = 18 numbers. S=18: 1 pair × 2 C = 2 numbers. Other S (1..8,10..17): total pairs = 90-9-1=80 pairs, each ×1 C = 80 numbers. Total = 18+2+80 = 100.
Coordinate geometry turns messy geometry into manageable algebra. Use it liberally.
Easier: Use generating functions or casework on positions of 4’s and 2/6’s. This is long — but the known answer from past solutions is . S=18: 1 pair × 2 C = 2 numbers
(x−258)2+(y−6)2=(258)2open paren x minus 25 over 8 end-fraction close paren squared plus open paren y minus 6 close paren squared equals open paren 25 over 8 end-fraction close paren squared The circle intersects side BCcap B cap C BCcap B cap C lies entirely on the x-axis, where . To find the x-coordinate of point , substitute into the circle's equation:
Solving National Sprint Round problems requires a shift in mindset from "How do I calculate this?" to "How does the author intend for me to solve this?"
-1m≡-4(mod9)negative 1 m triple bar negative 4 space open paren mod space 9 close paren 10..17): total pairs = 90-9-1=80 pairs
Note: These are sample-style problems that reflect Sprint characteristics; each solution focuses on the key insight and an efficient path rather than lengthy exposition.
r=a+b−c2r equals the fraction with numerator a plus b minus c and denominator 2 end-fraction Substitute our given lengths into this formula: