Difference between revisions of "1967 AHSME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | <math>\ | + | The area <math>K</math> of the triangle can be expressed in terms of its inradius <math>r</math> and its semiperimeter <math>s</math> as: |
| + | <cmath> K = r \times s = r \times \frac{P}{2}</cmath> | ||
| + | |||
| + | So, <math>\frac{P}{K} = \boxed{\textbf{(C) } \frac{2}{r}}</math>. | ||
| + | |||
| + | ~ proloto | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1967|num-b=4|num-a=6}} | + | {{AHSME 40p box|year=1967|num-b=4|num-a=6}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:33, 28 September 2023
Problem
A triangle is circumscribed about a circle of radius 
inches. If the perimeter of the triangle is 
inches and the area is 
square inches, then 
is:
Solution
The area of the triangle can be expressed in terms of its inradius and its semiperimeter 
as:
![\[K = r \times s = r \times \frac{P}{2}\]](http://latex.artofproblemsolving.com/6/b/2/6b2fab40f5c4b4a8f4917ac19fb6f7a29a92a0b7.png)
So, 
.
~ proloto
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
