Difference between revisions of "1958 AHSME Problems/Problem 43"
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import geometry; | import geometry; | ||
unitsize(50); | unitsize(50); | ||
− | pair | + | pair C = (0,0), B = (3,0), A = (0, 4); |
− | pair | + | pair AC = midpoint(A--C), BC = midpoint(B--C); |
draw(A--B--C--A); | draw(A--B--C--A); | ||
− | draw(A-- | + | draw(A--C, StickIntervalMarker(2, 1)); |
− | draw( | + | draw(B--C, StickIntervalMarker(2, 2)); |
− | |||
draw(B--AC); | draw(B--AC); | ||
− | + | draw(A--BC); | |
dot(AC); | dot(AC); | ||
+ | dot(BC); | ||
MP("$A$", A, W); | MP("$A$", A, W); | ||
MP("$B$", B, E); | MP("$B$", B, E); | ||
MP("$C$", C, W); | MP("$C$", C, W); | ||
− | MP("$ | + | MP("$E$", AC, W); |
− | + | MP("$D$", BC, S); | |
− | + | label("$y$", A--AC, W); | |
− | |||
− | label("$y$", A--AC, | ||
label("$y$", AC--C, W); | label("$y$", AC--C, W); | ||
− | draw(rightanglemark(C | + | label("$x$", B--BC, S); |
+ | label("$x$", BC--C, S); | ||
+ | draw(rightanglemark(A, C, B)); | ||
</asy> | </asy> | ||
− | <math>\fbox{D}</math> | + | |
+ | By the Pythagorean Theorem, <math>(2x)^2+y^2=BE^2</math>, and <math>x^2+(2y)^2=AD^2</math>. | ||
+ | |||
+ | Plugging in, <math>4x^2+y^2=16</math>, and <math>x^2+4y^2=49</math>. | ||
+ | |||
+ | Adding the equations, <math>5x^2+5y^2=65</math>, and dividing by <math>5</math>, <math>x^2+y^2=13</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Note that the length <math>AB^2</math> is equal to <math>(2x)^2+(2y)^2=4x^2+4y^2=4(x^2+y^2)</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\sqrt{4(13)}=2\sqrt{13}</math> <math>\fbox{D}</math> | ||
== See Also == | == See Also == |
Latest revision as of 12:46, 23 February 2018
Problem
is the hypotenuse of a right triangle
. Median
has length
and median
has length
. The length of
is:
Solution
By the Pythagorean Theorem, , and
.
Plugging in, , and
.
Adding the equations, , and dividing by
,
.
Note that the length is equal to
.
Therefore, the answer is
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 42 |
Followed by Problem 44 | |
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.