Difference between revisions of "1959 AHSME Problems/Problem 2"

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==Solution==
 
==Solution==
<math>\boxed{\textbf{D}}</math>
 
  
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<asy>
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import geometry;
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point A=(0,0);
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point B=(10,0);
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point C=(4,6);
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point P=(3,6(-sqrt(2)/2+1));
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point D,E;
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// Triangle ABC
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draw(A--B--C--A);
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dot(A);
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label("A",A,SW);
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dot(B);
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label("B",B,SE);
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dot(C);
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label("C",C,NW);
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// Point P and triangle CDE
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dot(P);
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label("P",P,N);
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pair[] d=intersectionpoints(parallel(P,line(A,B)),(B--C));
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D=d[0];
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dot(D);
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label("D",D,NE);
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pair[] e=intersectionpoints(parallel(P,line(A,B)),(A--C));
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E=e[0];
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dot(E);
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label("E",E,NW);
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draw(D--E);
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</asy>
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<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 11:50, 21 July 2024

Problem 2

Through a point $P$ inside the $\triangle ABC$ a line is drawn parallel to the base $AB$, dividing the triangle into two equal areas. If the altitude to $AB$ has a length of $1$, then the distance from $P$ to $AB$ is:

$\textbf{(A)}\ \frac12 \qquad\textbf{(B)}\ \frac14\qquad\textbf{(C)}\ 2-\sqrt2\qquad\textbf{(D)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(E)}\ \frac{2+\sqrt2}{8}$


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]

$\boxed{\textbf{(D)}}$

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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