Difference between revisions of "2000 AMC 10 Problems/Problem 19"

Latest revision as of 23:57, 26 November 2011

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-==Problem==
+#REDIRECT [[2000 AMC 12 Problems/Problem 21]]
-Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is <math>m</math> times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
-<math>\mathrm{(A)}\ \frac{1}{2m+1} \qquad\mathrm{(B)}\ m \qquad\mathrm{(C)}\ 1-m \qquad\mathrm{(D)}\ \frac{1}{4m} \qquad\mathrm{(E)}\ \frac{1}{8m^2}</math>
-==Solution==
-==See Also==
-{{AMC10 box|year=2000|num-b=18|num-a=20}}