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Difference between revisions of "1958 AHSME Problems/Problem 43"

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import geometry;
 
import geometry;
 
unitsize(50);
 
unitsize(50);
−
pair A = (0,0), B = (3,0), C = (0, 4);
+
pair C = (0,0), B = (3,0), A = (0, 4);
−
pair AB = midpoint(A--B), AC = midpoint(A--C);
+
pair AC = midpoint(A--C), BC = midpoint(B--C);
 
draw(A--B--C--A);
 
draw(A--B--C--A);
−
draw(A--B, StickIntervalMarker(2, 1));
+
draw(A--C, StickIntervalMarker(2, 1));
−
draw(A--C, StickIntervalMarker(2, 2));
+
draw(B--C, StickIntervalMarker(2, 2));
−
draw(C--AB);
 
 
draw(B--AC);
 
draw(B--AC);
−
dot(AB);
+
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("$M$", AB, S);
+
MP("$E$", AC, W);
−
MP("$N$", AC, W);
+
MP("$D$", BC, S);
−
label("$x$", A--AB, S);
+
label("$y$", A--AC, W);
−
label("$x$", AB--B, S);
 
−
label("$y$", A--AC, SWW);
 
 
label("$y$", AC--C, W);
 
label("$y$", AC--C, W);
−
draw(rightanglemark(C, A, B));
+
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 11:46, 23 February 2018

Problem

$\overline{AB}$ is the hypotenuse of a right triangle $ABC$. Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$. The length of $\overline{AB}$ is:

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 5\sqrt{3}\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{13}\qquad \textbf{(E)}\ 2\sqrt{15}$

Solution

[asy] import geometry; unitsize(50); pair C = (0,0), B = (3,0), A = (0, 4); pair AC = midpoint(A--C), BC = midpoint(B--C); draw(A--B--C--A); draw(A--C, StickIntervalMarker(2, 1)); draw(B--C, StickIntervalMarker(2, 2)); draw(B--AC); draw(A--BC); dot(AC); dot(BC); MP("$A$", A, W); MP("$B$", B, E); MP("$C$", C, W); MP("$E$", AC, W); MP("$D$", BC, S); label("$y$", A--AC, W); label("$y$", AC--C, W); label("$x$", B--BC, S); label("$x$", BC--C, S); draw(rightanglemark(A, C, B)); [/asy]

By the Pythagorean Theorem, $(2x)^2+y^2=BE^2$, and $x^2+(2y)^2=AD^2$.

Plugging in, $4x^2+y^2=16$, and $x^2+4y^2=49$.

Adding the equations, $5x^2+5y^2=65$, and dividing by $5$, $x^2+y^2=13$.


Note that the length $AB^2$ is equal to $(2x)^2+(2y)^2=4x^2+4y^2=4(x^2+y^2)$.

Therefore, the answer is $\sqrt{4(13)}=2\sqrt{13}$ $\fbox{D}$

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

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