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Difference between revisions of "1958 AHSME Problems/Problem 19"

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== Problem ==
 
== Problem ==
  
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The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b \equal{} 1 : 3</math>, then the ratio of <math> r</math> to <math> s</math> is:
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The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b = 1 : 3</math>, then the ratio of <math> r</math> to <math> s</math> is:
  
 
<math> \textbf{(A)}\ 1 : 3\qquad  
 
<math> \textbf{(A)}\ 1 : 3\qquad  
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== Solution ==
 
== Solution ==
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Let the triangle be triangle <math>ABC</math> with <math>A</math> opposite to side <math>a</math>, and define <math>B</math> and <math>C</math> similarly. Call the base of the perpendicular <math>D</math> and say it has length <math>x</math>.
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We know the area of triangle <math>ABC</math>, denoted as <math>[ABC],</math> is <math>\frac{ab}{2} = \frac{cx}{2} = \frac{(r+s)x}{2}.</math> So, <math>x = \frac{ab}{r+s}.</math>
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Now, by simple angle chasing that <math>\triangle BCD \sim \triangle CDA \sim ACB.</math> So, <cmath>\frac{AC}{CB} = \frac{CD}{BD}.</cmath> Plugging in our variables for the side lenghts: <cmath>\frac{b}{a} = \frac{x}{r} = \frac{\frac{ab}{r+s}}{r}.</cmath>
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We now get that <math>br = \frac{a^2b}{r+s},</math> so <math>r = \frac{a^2}{r+s}.</math> Similarly, we get that <math>s = \frac{b^2}{r+s}.</math> So, <cmath>\frac{r}{s} =  \frac{\frac{a^2}{r+s}}{\frac{b^2}{r+s}} = \frac{a^2}{b^2} = \boxed{\textbf{(B)}\ 1 : 9}</cmath>
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<math>\fbox{}</math>
 
<math>\fbox{}</math>
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~Makethan
  
 
== See Also ==
 
== See Also ==

Latest revision as of 01:54, 21 November 2024

Problem

The sides of a right triangle are $a$ and $b$ and the hypotenuse is $c$. A perpendicular from the vertex divides $c$ into segments $r$ and $s$, adjacent respectively to $a$ and $b$. If $a : b = 1 : 3$, then the ratio of $r$ to $s$ is:

$\textbf{(A)}\ 1 : 3\qquad \textbf{(B)}\ 1 : 9\qquad \textbf{(C)}\ 1 : 10\qquad \textbf{(D)}\ 3 : 10\qquad \textbf{(E)}\ 1 : \sqrt{10}$

Solution

Let the triangle be triangle $ABC$ with $A$ opposite to side $a$, and define $B$ and $C$ similarly. Call the base of the perpendicular $D$ and say it has length $x$.

We know the area of triangle $ABC$, denoted as $[ABC],$ is $\frac{ab}{2} = \frac{cx}{2} = \frac{(r+s)x}{2}.$ So, $x = \frac{ab}{r+s}.$

Now, by simple angle chasing that $\triangle BCD \sim \triangle CDA \sim ACB.$ So, \[\frac{AC}{CB} = \frac{CD}{BD}.\] Plugging in our variables for the side lenghts: \[\frac{b}{a} = \frac{x}{r} = \frac{\frac{ab}{r+s}}{r}.\]

We now get that $br = \frac{a^2b}{r+s},$ so $r = \frac{a^2}{r+s}.$ Similarly, we get that $s = \frac{b^2}{r+s}.$ So, \[\frac{r}{s} = \frac{\frac{a^2}{r+s}}{\frac{b^2}{r+s}} = \frac{a^2}{b^2} = \boxed{\textbf{(B)}\ 1 : 9}\]

$\fbox{}$

~Makethan

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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