MathDB

14

Part of 2010 AMC 10

Problems(2)

Triangle with AB = 2*AC

Source: AMC 12 2010A, Problem 8

2/10/2010
Triangle ABC ABC has AB \equal{} 2 \cdot AC. Let D D and E E be on AB‾ \overline{AB} and BC‾ \overline{BC}, respectively, such that \angle{BAE} \equal{} \angle{ACD}. Let F F be the intersection of segments AE AE and CD CD, and suppose that △CFE \triangle{CFE} is equilateral. What is ∠ACB \angle{ACB}? <spanclass=′latex−bold′>(A)</span> 60∘<spanclass=′latex−bold′>(B)</span> 75∘<spanclass=′latex−bold′>(C)</span> 90∘<spanclass=′latex−bold′>(D)</span> 105∘<spanclass=′latex−bold′>(E)</span> 120∘ <span class='latex-bold'>(A)</span>\ 60^{\circ}\qquad <span class='latex-bold'>(B)</span>\ 75^{\circ}\qquad <span class='latex-bold'>(C)</span>\ 90^{\circ}\qquad <span class='latex-bold'>(D)</span>\ 105^{\circ}\qquad <span class='latex-bold'>(E)</span>\ 120^{\circ}
trigonometrygeometryratiotrig identitiesLaw of Sinesexterior angleAMC 10 2010 number 14
Averages

Source: 2010 AMC 10B Problem 14

2/25/2010
The average of the numbers 1,2,3,...,98,99 1,2,3,...,98,99, and x x is 100x 100x. What is x x? <spanclass=′latex−bold′>(A)</span> 49101<spanclass=′latex−bold′>(B)</span> 50101<spanclass=′latex−bold′>(C)</span> 12<spanclass=′latex−bold′>(D)</span> 51101<spanclass=′latex−bold′>(E)</span> 5099 <span class='latex-bold'>(A)</span>\ \frac{49}{101} \qquad<span class='latex-bold'>(B)</span>\ \frac{50}{101} \qquad<span class='latex-bold'>(C)</span>\ \frac12 \qquad<span class='latex-bold'>(D)</span>\ \frac{51}{101} \qquad<span class='latex-bold'>(E)</span>\ \frac{50}{99}
AMC