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Difference between revisions of "1986 AHSME Problems/Problem 11"

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==FATSO==
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==Problem==
  
 
In <math>\triangle ABC, AB = 13, BC = 14</math> and <math>CA = 15</math>. Also, <math>M</math> is the midpoint of side <math>AB</math> and <math>H</math> is the foot of the altitude from <math>A</math> to <math>BC</math>.  
 
In <math>\triangle ABC, AB = 13, BC = 14</math> and <math>CA = 15</math>. Also, <math>M</math> is the midpoint of side <math>AB</math> and <math>H</math> is the foot of the altitude from <math>A</math> to <math>BC</math>.  
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In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is <math>13</math>, the median must be <math>6.5</math>.
 
In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is <math>13</math>, the median must be <math>6.5</math>.
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== Solution 2 (Self Torture) ==
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Warning: this solution is very intensive in calculation. Please do NOT try this on the test!
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Let's start by finding <math>AH</math>. By Heron's Formula, <math>s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84</math>. Using the area formula <math>A=0.5bh</math>, <math>AH=12</math>. Now using the Pythagorean Theorem, <math>BH=5, HC=9</math>.
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Now <math>AM=MB=6.5</math>. Using Stewart's Theorem on <math>\triangle{ABH}</math>, letting <math>HM=x</math>:
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<math>13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5</math> (remember that Stewart's Theorem is <math>man+dad=bmb+cnc</math>).
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Thus <math>x=6.5</math> or <math>x=-6.5</math> (reject this solution since <math>x</math> is positive). Thus <math>HM=6.5</math>. Select <math>\boxed{B}</math>.
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~hastapasta
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P.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.
  
 
== See also ==
 
== See also ==

Latest revision as of 10:41, 1 December 2022

Problem

In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$. Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$. The length of $HM$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$

Solution

In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is $13$, the median must be $6.5$.

Solution 2 (Self Torture)

Warning: this solution is very intensive in calculation. Please do NOT try this on the test!

Let's start by finding $AH$. By Heron's Formula, $s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84$. Using the area formula $A=0.5bh$, $AH=12$. Now using the Pythagorean Theorem, $BH=5, HC=9$.

Now $AM=MB=6.5$. Using Stewart's Theorem on $\triangle{ABH}$, letting $HM=x$:

$13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5$ (remember that Stewart's Theorem is $man+dad=bmb+cnc$).

Thus $x=6.5$ or $x=-6.5$ (reject this solution since $x$ is positive). Thus $HM=6.5$. Select $\boxed{B}$.

~hastapasta

P.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
All AHSME Problems and Solutions

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