Difference between revisions of "1995 AHSME Problems/Problem 23"

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== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
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{{MAA Notice}}

Latest revision as of 13:05, 5 July 2013

Problem

The sides of a triangle have lengths $11,15,$ and $k$, where $k$ is an integer. For how many values of $k$ is the triangle obtuse?

$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 13 } \qquad \mathrm{(E) \ 14 }$

Solution

By the Law of Cosines, a triangle is obtuse if the sum of the squares of two of the sides of the triangles is less than the square of the third. The largest angle is either opposite side $15$ or side $k$. If $15$ is the largest side,

\[15^2 >11^2 + k^2 \Longrightarrow k < \sqrt{104}\]

By the Triangle Inequality we also have that $k > 4$, so $k$ can be $5, 6, 7, \ldots , 10$, or $6$ values.

If $k$ is the largest side,

\[k^2 >11^2 + 15^2 \Longrightarrow k > \sqrt{346}\]

Combining with the Triangle Inequality $19 \le k < 26$, or $7$ values. These total $13\ \mathrm{(D)}$ values of $k$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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
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