Difference between revisions of "2002 AMC 12P Problems/Problem 16"
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== Solution == | == Solution == | ||
| − | + | Let <math>a, b,</math> and <math>c</math> denote the bases of altitudes <math>12, 15,</math> and <math>20,</math> respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so <math>\frac{12a}{2}=\frac{15b}{2}=\frac{20c}{2}.</math> Multiplying by <math>2</math>, we get <math>12a=15b=20c.</math> Notice that a simple solution to the equation is if all of them equal <math>12 \cdot 15 \cdot 20.</math> That means <math>a=15 \cdot 20</math>, b=12 \cdot 20,<math> and </math>c=12 \cdot 15.$ Simplifying, we see that | |
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=15|num-a=17}} | {{AMC12 box|year=2002|ab=P|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:51, 31 December 2023
Problem
The altitudes of a triangle are 
and 
The largest angle in this triangle is
Solution
Let 
and 
denote the bases of altitudes and 
respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so 
Multiplying by 
, we get 
Notice that a simple solution to the equation is if all of them equal 
That means 
, b=12 \cdot 20,
c=12 \cdot 15.$ Simplifying, we see that
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
