Difference between revisions of "1959 AHSME Problems/Problem 2"
Treetor10145 (talk | contribs) (Added Problem, Solution, and See Also) |
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==Solution== | ==Solution== | ||
− | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A=(0,0); | ||
+ | point B=(10,0); | ||
+ | point C=(4,6); | ||
+ | point P=(3,6(-sqrt(2)/2+1)); | ||
+ | point D,E; | ||
+ | |||
+ | // Triangle ABC | ||
+ | draw(A--B--C--A); | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("B",B,SE); | ||
+ | dot(C); | ||
+ | label("C",C,NW); | ||
+ | |||
+ | // Point P and triangle CDE | ||
+ | dot(P); | ||
+ | label("P",P,N); | ||
+ | |||
+ | pair[] d=intersectionpoints(parallel(P,line(A,B)),(B--C)); | ||
+ | D=d[0]; | ||
+ | dot(D); | ||
+ | label("D",D,NE); | ||
+ | |||
+ | pair[] e=intersectionpoints(parallel(P,line(A,B)),(A--C)); | ||
+ | E=e[0]; | ||
+ | dot(E); | ||
+ | label("E",E,NW); | ||
+ | |||
+ | draw(D--E); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | <math>\boxed{\textbf{(D)}}</math> | ||
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=1|num-a=3}} | {{AHSME 50p box|year=1959|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:50, 21 July 2024
Problem 2
Through a point inside the a line is drawn parallel to the base , dividing the triangle into two equal areas. If the altitude to has a length of , then the distance from to is:
Solution
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.