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Difference between revisions of "2002 AMC 12P Problems/Problem 16"

(Created page with "== Problem == How many positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B...")
 
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== Problem ==
 
== Problem ==
−
How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+
The altitudes of a triangle are <math>12, 15,</math> and <math>20.</math> The largest angle in this triangle is
  
−
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }  </math>
+
<math>
 +
\text{(A) }72^\circ
 +
\qquad
 +
\text{(B) }75^\circ
 +
\qquad
 +
\text{(C) }90^\circ
 +
\qquad
 +
\text{(D) }108^\circ
 +
\qquad
 +
\text{(E) }120^\circ
 +
</math>
  
 
== Solution ==
 
== Solution ==
−
If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
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, b=12 \cdot 20,</math> and <math>c=12 \cdot 15.</math> Simplifying our solution to check for Pythagorean triples we see that this is just a Pythagorean triple, namely a <math>3-4-5</math> triangle. Since the other two angles of a right triangle must be acute, the right angle must be the greatest angle. Therefore, our answer is <math>\boxed{\textbf{(C) }90^\circ}.</math>
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC12 box|year=2002|ab=P|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:55, 31 December 2023

Problem

The altitudes of a triangle are $12, 15,$ and $20.$ The largest angle in this triangle is

$\text{(A) }72^\circ \qquad \text{(B) }75^\circ \qquad \text{(C) }90^\circ \qquad \text{(D) }108^\circ \qquad \text{(E) }120^\circ$

Solution

Let $a, b,$ and $c$ denote the bases of altitudes $12, 15,$ and $20,$ respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so $\frac{12a}{2}=\frac{15b}{2}=\frac{20c}{2}.$ Multiplying by $2$, we get $12a=15b=20c.$ Notice that a simple solution to the equation is if all of them equal $12 \cdot 15 \cdot 20.$ That means $a=15 \cdot 20, b=12 \cdot 20,$ and $c=12 \cdot 15.$ Simplifying our solution to check for Pythagorean triples we see that this is just a Pythagorean triple, namely a $3-4-5$ triangle. Since the other two angles of a right triangle must be acute, the right angle must be the greatest angle. Therefore, our answer is $\boxed{\textbf{(C) }90^\circ}.$

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

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