1964 AHSME Problems/Problem 35

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 $\dfrac{3.75}{8.25}=\dfrac{\frac{15}{4}}{\frac{33}{4}}=\dfrac{15}{33}=\dfrac{5}{11}=\boxed{\bold{B}}.$ -OreoChocolate

Solution 3 (Ceva's Theorem)

$[asy] size(150); real a = 14, b = 15, c = 13; pair C = (0, 0), B = (a, 0); // calculate cos(α) and sin(α) real cos_alpha = (a^2 + c^2 - b^2) / (2 * a * c); real sin_alpha = sqrt(1 - cos_alpha^2); // calculate coordinates of A pair A = (c * cos_alpha, c * sin_alpha); // calculate altitudes pair D = foot(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); // calculate orthocenter pair H = extension(A, D, B, E); // draw triangle and altitudes draw(A--B--C--cycle); draw(A--D, dashed); draw(B--E, dashed); draw(C--F, dashed); // draw right angle markers draw(rightanglemark(A, D, B, 20)); draw(rightanglemark(B, E, A, 20)); draw(rightanglemark(C, F, A, 20)); // label points dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, NW); dot("$F$", F, NE); dot("$H$", H, SE); // label side lengths label("$13$", (A+C)/2, NW); label("$15$", (A+B)/2, NE); label("$14$", (B+C)/2, S); [/asy]$

A consequence of Ceva's Theorem that is sometimes attributed to van Aubel states that:

$\dfrac{AH}{HD} = \dfrac{AE}{EC} + \dfrac{AF}{FB}$

We must then find $AE$ and $AF$ . To find $AF$ note that $\triangle AFC$ and $\triangle BFC$ are both right triangles sharing a common height, $FC$ . Thus

$AF^2+FC^2=AC^2, \text{and } BF^2+FC^2=BC^2 \implies$ $AF^2+FC^2=13^2, \text{and } (15-AF)^2+FC^2=14^2$

Subtracting the two equations to eliminate the common height term ( $FC^2$ ):

$(15-AF)^2-AF^2=27 \implies AF=\dfrac{33}{5}$

A similar computation using $\triangle AEB$ and $\triangle CEB$ gives us:

$AE^2+EB^2=AB^2, \text{and } CE^2+EB^2=BC^2 \implies$ $AE^2+EB^2=15^2, \text{and } (13-AE)^2+EB^2=14^2$

$AE^2-(13-AE)^2=29 \implies AE=\dfrac{99}{13}$

Returning to our original van Aubel equation yields:

$\dfrac{AH}{HD} = \dfrac{AE}{EC} + \dfrac{AF}{FB} = \dfrac{\dfrac{99}{13}}{13-\dfrac{99}{13}} + \dfrac{\dfrac{33}{5}}{15-\dfrac{33}{5}}=\dfrac{99}{70}+\dfrac{55}{70}=\dfrac{11}{5}$

Therefore $HD:HA = \boxed{\textbf{(B) }5:11}$

~ proloto

Solution 4 (Similar Triangles)

Because this triangle is a nice triangle, we can easily find that $AD = 12$ , $CD = 5$ , and $DB = 9$ . From this, we know that: $\triangle BDA$ is a 3:4:5 triangle, and it is similar to $\triangle BEC$ which is similar to $\triangle CDH$ . Therefore $HD = CD \cdot 3/4 = 15/4$ , and $AH = 12-15/4 = 33/4$ , $HD/AH = \dfrac{15/4}{33/4} = \dfrac{5}{11}=\boxed{\bold{B}}$ . -ELIZABETHWZYING

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 36
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