MathDB

23

Part of 2002 AMC 10

Problems(3)

Triangle

Source: AMC 2002 10A #23

2/11/2008
Points A,B,C A,B,C and D D lie on a line, in that order, with AB\equal{}CD and BC\equal{}12. Point E E is not on the line, and BE\equal{}CE\equal{}10. The perimeter of △AED \triangle AED is twice the perimeter of △BEC \triangle BEC. Find AB AB. (A) 15/2(B) 8(C) 17/2(D) 9(E) 19/2 \text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2
geometryperimeterAMC
Sequence

Source:

2/27/2008
Let {ak} \{a_k\} be a sequence of integers such that a_1 \equal{} 1 and a_{m \plus{} n} \equal{} a_m \plus{} a_n \plus{} mn, for all positive integers m m and n n. Then a12 a_{12} is <spanclass=′latex−bold′>(A)</span> 45<spanclass=′latex−bold′>(B)</span> 56<spanclass=′latex−bold′>(C)</span> 67<spanclass=′latex−bold′>(D)</span> 78<spanclass=′latex−bold′>(E)</span> 89 <span class='latex-bold'>(A)</span>\ 45 \qquad <span class='latex-bold'>(B)</span>\ 56 \qquad <span class='latex-bold'>(C)</span>\ 67 \qquad <span class='latex-bold'>(D)</span>\ 78 \qquad <span class='latex-bold'>(E)</span>\ 89
functioninduction
Differences in Values

Source:

4/2/2013
Let a=121+223+325+⋯+100122001a=\dfrac{1^2}1+\dfrac{2^2}3+\dfrac{3^2}5+\cdots+\dfrac{1001^2}{2001} and b=123+225+327+⋯+100122003.b=\dfrac{1^2}3+\dfrac{2^2}5+\dfrac{3^2}7+\cdots+\dfrac{1001^2}{2003}. Find the integer closest to a−ba-b.
<spanclass=′latex−bold′>(A)</span>500<spanclass=′latex−bold′>(B)</span>501<spanclass=′latex−bold′>(C)</span>999<spanclass=′latex−bold′>(D)</span>1000<spanclass=′latex−bold′>(E)</span>1001<span class='latex-bold'>(A) </span>500\qquad<span class='latex-bold'>(B) </span>501\qquad<span class='latex-bold'>(C) </span>999\qquad<span class='latex-bold'>(D) </span>1000\qquad<span class='latex-bold'>(E) </span>1001