Difference between revisions of "1959 AHSME Problems/Problem 2"
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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> | ||
| + | |||
| + | Because <math>[ABDE]=[\triangle EDC]</math>, we know that <math>[\triangle EDC]=\frac{1}{2}[\triangle ABC]</math>. Because the ratio of their areas is <math>\frac{1}{2}</math>, we know that the ratio of their side lengths (and thus their altitudes) is <math>\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}</math>. Thus, the altitude from <math>C</math> to <math>\overline{DE}</math> has length <math>1*\frac{\sqrt2}{2}=\frac{\sqrt{2}}{2}</math>. Thus, because <math>\overline{DE} \parallel \overline{AB}</math>, the distance from <math>P</math> to <math>\overline{AB}</math> is <math>1-\frac{\sqrt{2}}{2}=\boxed{\textbf{(D) }\frac{2-\sqrt{2}}{2}}</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}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 10:59, 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
![[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]](http://latex.artofproblemsolving.com/e/6/d/e6d8d89db467e374f39ce656c99118c31fb0cea5.png)
Because ![$[ABDE]=[\triangle EDC]$](http://latex.artofproblemsolving.com/2/c/8/2c80d6aea325b78a3e76be42f373931ed28543fd.png)
, we know that ![$[\triangle EDC]=\frac{1}{2}[\triangle ABC]$](http://latex.artofproblemsolving.com/3/2/e/32ecf884e3cb6d5f2f2a6669047a5d4f1fd31bae.png)
. Because the ratio of their areas is 
, we know that the ratio of their side lengths (and thus their altitudes) is 
. Thus, the altitude from 
to 
has length 
. Thus, because 
, the distance from to 
is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
