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1964 AHSME Problems/Problem 35

Revision as of 13:19, 21 April 2020 by Oreochocolate (talk | contribs) (added another possibly time-consuming but good solution, i guess)

Problem

The sides of a triangle are of lengths $13$, $14$, and $15$. The altitudes of the triangle meet at point $H$. if $AD$ is the altitude to the side of length $14$, the ratio $HD:HA$ is:

$\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33$

Solution 1

Using Law of Cosines and the fact that the ratio equals cos(a)/[cos(b)cos(c)] B 5:11

Solution 2 (coordinates)

[asy] draw((0,0)--(15,0)--(6.6,11.2)--(0,0)); draw((0,0)--(9.6,7.2)); draw((6.6,0)--(6.6,11.2)); draw((15,0)--(3267/845,5544/845)); label("$B$",(15,0),SE); label("$C$",(6.6,11.2),N); label("$E$",(6.6,0),S); label("$15$",(7.5,-0.75),S); label("$14$",(11,5.75),ENE); label("$13$",(3,6),WNW); label("$A$",(0,0),SW); label("$D$",(9.6,7.2),NE); label("$H$",(6.6,3.5),E); [/asy] The reason why we have $HD$ shorter than $HA$ is that all the ratios' left hand side ($HD$) is less than the ratios' right hand side ($HA$).

We label point $A$ as the origin and point $B$, logically, as $(15,0)$. By Heron's Formula, the area of this triangle is $84.$ Thus the height perpendicular to $AB$ has a length of $11.2,$ and by the Pythagorean Theorem, $AE$ and $EB$ have lengths $6.6$ and $8.4,$ respectively. These lengths tell us that $C$ is at $(6.6,11.2)$.

The slope of $BC$ is $\dfrac{0-11.2}{15-6.6}=-\dfrac{4}{3},$ and the slope of $AD$ is $\dfrac{3}{4}$ by taking the negative reciprocal of $-\dfrac{4}{3}.$ Therefore, the equation of line $AD$ can best be represented by $y=\dfrac{3}{4}x.$

We next find the intersection of $CE$ and $AD$. We automatically know the $x$-value; it is just $6.6$ because $CE$ is a straight line hitting $(6.6,0).$ Therefore, the $y$-value is at $\dfrac{3}{4}\times 6.6=4.95.$ Therefore, the intersection between $CE$ and $AD$ is at $(6.6,4.95)$.

We also need to find the intersection between $BC$ and $AD$. To do that, we know that the line of $AD$ is represented as $y=\dfrac{3}{4}x,$ and the slope of line $BC$ is $-\dfrac{4}{3}.$ We just need to find line $BC$'s y-intercept. So far, we have $y=-\dfrac{4}{3}x+b,$ where $b$ is a real y-intercept. We know that $B$ is located at $(15,0),$ so we plug that into the equation and yield $b=20.$ Therefore, the intersection between the two lines is \[\dfrac{3}{4}x=-\dfrac{4}{3}x+20, 9x=-16x+240, 25x=240, x=9.6, y=\dfrac{3}{4}\times 9.6, y=7.2.\] After that, we use the distance formula: $HA$ has a length of \[\sqrt{(6.6-0)^2+(4.95-0)^2}=\sqrt{\dfrac{1089}{25}+\dfrac{9801}{400}}=\sqrt{\dfrac{1089*16+9801}{400}}=\sqrt{\dfrac{27225}{400}}=\sqrt{\dfrac{1089}{16}}=\dfrac{33}{4}=8.25,\] and $HD$ has a length of \[\sqrt{(9.6-6.6)^2+(7.2-4.95)^2}=\sqrt{3^2+(\dfrac{36}{5}-\dfrac{99}{20})^2}=\sqrt{9+\dfrac{81}{16}}=\sqrt{\dfrac{225}{16}}=3.75.\] Thus, we have that $\dfrac{3.75}{8.25}=\dfrac{\frac{15}{4}}{\frac{33}{4}}=\dfrac{15}{33}=\dfrac{5}{11}=\bold{\boxed{B}}.$

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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
All AHSME Problems and Solutions

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