Patricia builds two triangles, each with 30 feet of wood. The first triangle ABC is built to maximize the length of the base side. The second triangle DEF is built to maximize the area of the triangle. What is the ratio of the length of the base of triangle ABC to the length of the base of triangle DEF? The lengths of all line segments are integers.

A. 1:1
B. 6:5
C. 5:4
D. 7:5
E. 2:1
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The lengths of 3 sides of triangle ABC is a, b, c where a, b, c are integers and c is the base side.
We have \(c<a+b \implies 2c < a+b+c=30 \implies c<15\). Triangle ABC is built to maximize \(c\), so \(c=14\). The length of base side is 14.

For any triangle that has the same perimeter, equilateral triangle has maximum area. So triangle DEF has 3 sides \(d=e=f=10\). The length of base side is 10.

Hence, the ratio is 14 : 10 = 7 : 5

The answer is D.
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I use Heron's formula:

\(S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)}\) where \(p=\frac{a+b+c}{2}=const\)

From inequation: \(xyz \leq ( \frac{x+y+z}{3})^3\) for \(x, y, z>0\). Sign \("="\) occurs \(\iff x=y=z\)

We have \(S_{ABC} \leq \sqrt{p \times \left ( \frac{p-a+p-b+p-c}{3} \right) ^3}= \sqrt{p \times (\frac{p}{3})^3}=\frac{p^2}{3\sqrt{3}}\)

\(maxS_{ABC}=\frac{p^2}{3\sqrt{3}} \iff a=b=c=\frac{2p}{3}\)
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c < a + b => c + c < a + b+ c => 2c < a + b + c
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Hi All,

We're told that Patricia builds two triangles, EACH with 30 feet of wood. The first triangle ABC is built to maximize the LENGTH of the BASE side. The second triangle DEF is built to maximize the AREA of the triangle. In addition, the lengths of all line segments are INTEGERS. We're asked for the ratio of the length of the base of triangle ABC to the length of the base of triangle DEF. This question has a number of built-in 'logic shortcuts' to it; when combined with the proper Geometry rules, you can actually answer this question without doing much 'math' at all.

To start, it's important to note that all of the side lengths MUST be INTEGERS - and the perimeters of each triangle are 30 - both of these facts limit the number of potential triangles that we can make.

For triangle ABC, we want the largest possible base, meaning that we'll be creating a really long, 'almost flat' triangle. Here, the Triangle Inequality Theorem will be useful. In simple terms, that math rule means that the sum of any two sides of a triangle MUST be greater than the third side... so the two sides that are NOT the base of ABC must sum to a total that is GREATER than the base. Since we're dealing with integers, the base CANNOT be 15... since that would make the sum of the other two sides 15... and 15 is NOT greater than 15. Thus, the base must be 14 and the sum of the other two sides would be 16.

Since the first part of the ratio is 14, we can eliminate Answers B and C (since they cannot be reduced from 14).

For triangle DEF, we need to find the base... but a triangle has 3 sides and any of them could be the base ... so which side (DE, DF or EF) would be the base? Notice how the answer choices all involve numbers (no variables), so the correct answer implies that the 'base' of DEF cannot possibly be 3 different values. What type of triangle does NOT have 3 different side lengths (or even 2 different side lengths for that matter)? An equilateral triangle! Thus, the three sides of DEF must be 10, 10 and 10. Mathematically-speaking, an equilateral triangle is how you would get the largest area, but that knowledge isn't necessary to answer this question.

The ratio of the two bases is 14:10 --> 7:5

Final Answer:

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\(x + y + z = 30\,\,\,\,\,\left( {\Delta \,\,{\rm{lengths}}\,\,{\rm{positive}}\,\,{\rm{ints}}\,\,{\rm{,}}\,\,\,x\,\,{\rm{base}}} \right)\)

\(?\,\,\, = \,\,\,\,{{x\,\,\,\left( {x\,\,\max } \right)} \over {x\,\,\,\,\left( {{S_\Delta }\max } \right)\,\,\,}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,{7 \over 5}\)

\(\left( * \right)\,\,\,\left\{ \matrix{
x\,\,\,\left( {x\,\,\max } \right)\,\,\,\,::\,\,\,\,\,\,x < y + z = 30 - x\,\,\,\,\,\left( {{\rm{the}}\,\,{\rm{giant}}\,\,{\rm{argument}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,x < 15\,\,\,\,\,\, \Rightarrow \,\,\,\,x\,\,\,\left( {x\,\,\max } \right) = 14\,\,\,\,\,\,\,\left[ {\left( {x,y,z} \right) = \left( {14,8,8} \right)\,\,\,{\rm{viable}}} \right] \hfill \cr
x\,\,\,\,\left( {{S_\Delta }\max } \right)\,\,\,::\,\,\,x = y = z = 10\,\,\,\,\left( {{\rm{perim}}\,\,{\rm{const,}}\,\,{\rm{max}}\,\,{\rm{area}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{regularity}}} \right) \hfill \cr} \right.\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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