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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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To find BD, we need to first find the altitude. If we have the angles of the triangle, we can use law of sines to find the length of the altitude. We know that the total degrees in a triangle is 180, so we can use law of sines to set up a proportion. The sum of the side lengths is 156. Diving 180 by 156 gives us each 15/13 for each side unit. Multiplying by 53, we get angle B is 795/13 degrees. Using law of sines gives us AD is about 44.67. Using pythagorean theorem, we find that BD is about 24.6.
 
  
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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.
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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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<cmath>
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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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</cmath>
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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 ==
 
{{UNCO Math Contest box|year=1993|n=II|num-b=9|after=Last Question}}
 
{{UNCO Math Contest box|year=1993|n=II|num-b=9|after=Last Question}}
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{{Heron's Formula}}
 
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 23:57, 3 June 2022

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)
Preceded by
Problem 9 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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