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Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 10"

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== Solution ==
 
== Solution ==
  
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Let BD=x. First, we notice that line segment AD is common to both right triangles ADB and ADC. We can therefore use the Pythagorean Theorem to say that (AD)^2 = 51^2 - x^2 = 53^2 - (52-x)^2. Solving for x, we get x=24.
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We will solve both parts at once since it is easy to get the two answers from this all-inclusive solution.
  
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To find the area, we simply use Heron's Formula to get 1170. (semi-perimeter is 78).
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Heron's Formula states that in a triangle with sides <math>a, b, c</math> and <math>s = \frac{a + b + c}{2},</math> the area is given by <cmath>\sqrt{s(s - a)(s - b)(s - c)}.</cmath> We plug in <math>a = 52, b = 53, c = 51.</math>
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\begin{align*}
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s &= \dfrac{51 + 52 + 53}{2} = \dfrac{3 \cdot 52}{2} = 3 \cdot 26 = 78 \\
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[ABC] &= \sqrt{78(78 - 51)(78 - 52)(78 - 53)} = \sqrt{78(27)(26)(25)} = 5 \sqrt{(3 \cdot 26)\left(3^3\right)(2 \cdot 13)} \\
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&= 5 \cdot 3 \sqrt{3 \cdot 2 \cdot 13 \cdot 3 \cdot 2 \cdot 13} = 15 \sqrt{2^2 \cdot 3^2 \cdot 13^2} = 15 \sqrt{(2 \cdot 3 \cdot 13)}^2 \\
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&= \boxed{1170}
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\end{align*}
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 +
Since <math>[ABC] = \frac{bh}{2},</math> we know that <math>1170 = \frac{AD \cdot 52}{2} = 26 \cdot AD.</math> Solving, we get <math>AD = 45.</math> Remembering our 8-15-17 Pythagorean triple, we see that <math>BD = \boxed{24}.</math> <math>\square</math>
  
 
== See also ==
 
== See also ==

Revision as of 15:04, 26 May 2017

Problem

The scalene triangle $ABC$ has side lengths $51, 52, 53.$ $AD$ is perpendicular to $BC.$ [asy] draw((0,0)--(52,0)--(24,sqrt(3)*26)--cycle); draw((24,0)--(24,sqrt(3)*26)); draw((0,-8)--(52,-8),arrow=Arrow()); draw((52,-8)--(0,-8),arrow=Arrow()); draw((24,3)--(21,3)--(21,0),black); MP("B",(0,0),SW);MP("A",(24,sqrt(3)*26),N);MP("C",(52,0),SE);MP("D",(24,0),S); MP("52",(26,-8),S);MP("53",(38,sqrt(3)*13),NE);MP("51",(12,sqrt(3)*13),NW); [/asy]

(a) Determine the length of $BD.$

(b) Determine the area of triangle $ABC.$


Solution

We will solve both parts at once since it is easy to get the two answers from this all-inclusive solution.

Heron's Formula states that in a triangle with sides $a, b, c$ and $s = \frac{a + b + c}{2},$ the area is given by \[\sqrt{s(s - a)(s - b)(s - c)}.\] We plug in $a = 52, b = 53, c = 51.$

\begin{align*} s &= \dfrac{51 + 52 + 53}{2} = \dfrac{3 \cdot 52}{2} = 3 \cdot 26 = 78 \\ [ABC] &= \sqrt{78(78 - 51)(78 - 52)(78 - 53)} = \sqrt{78(27)(26)(25)} = 5 \sqrt{(3 \cdot 26)\left(3^3\right)(2 \cdot 13)} \\ &= 5 \cdot 3 \sqrt{3 \cdot 2 \cdot 13 \cdot 3 \cdot 2 \cdot 13} = 15 \sqrt{2^2 \cdot 3^2 \cdot 13^2} = 15 \sqrt{(2 \cdot 3 \cdot 13)}^2 \\ &= \boxed{1170} \end{align*}

Since $[ABC] = \frac{bh}{2},$ we know that $1170 = \frac{AD \cdot 52}{2} = 26 \cdot AD.$ Solving, we get $AD = 45.$ Remembering our 8-15-17 Pythagorean triple, we see that $BD = \boxed{24}.$ $\square$

See also

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
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All UNCO Math Contest Problems and Solutions

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