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solid geometry - Triangle Inequality

For COMPETITION
Number of Total Problems: 5.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
The sides of a triangle with positive area have lengths $4$ , $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ , $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$ ?
$mathrm{(A)} 2 qquadmathrm{(B)} 4 qquadmathrm{(C)} 6 qquadmathrm{(D)} 8 qquadmathrm{(E)} 10$
'

Category Triangle Inequality

Analysis

Solution/Answer
From the triangle inequality, $2<x<10$ and $2<y<10$ . The smallest positive number not possible is $10-2$ , which is $8$ . $oxed{ ext{D}}$

Answer:

Problem Num : 2

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

How many non-congruent triangles with perimeter $7$ have integer side lengths?
$mathrm{(A) } 1qquad mathrm{(B) } 2qquad mathrm{(C) } 3qquad mathrm{(D) } 4qquad mathrm{(E) } 5$
''>''
How many non-congruent triangles with perimeter $7$ have integer side lengths?
$mathrm{(A) } 1qquad mathrm{(B) } 2qquad mathrm{(C) } 3qquad mathrm{(D) } 4qquad mathrm{(E) } 5$
''

Category Triangle Inequality

Analysis

Solution/Answer
By the triangle inequality, no side may have a length greater than the semiperimeter, which is $frac{1}{2}cdot7=3.5$ .
Since all sides must be integers, the largest possible length of a side is $3$ . Therefore, all such triangles must have all sides of length $1$ , $2$ , or $3$ . Since $2+2+2=6<7$ , at least one side must have a length of $3$ . Thus, the remaining two sides have a combined length of $7-3=4$ . So, the remaining sides must be either $3$ and $1$ or $2$ and $2$ . Therefore, the number of triangles is $oxed{mathrm{(B)} 2}$ .

Answer:

Problem Num : 3

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$mathrm{(A) } 43qquad mathrm{(B) } 44qquad mathrm{(C) } 45qquad mathrm{(D) } 46qquad mathrm{(E) } 47$
'

Category Triangle Inequality

Analysis

Solution/Answer
Let $x$ be the length of the first side.
The lengths of the sides are: $x$ , $3x$ , and $15$ .
By the Triangle Inequality,
$3x < x + 15$
$2x < 15$
$x < frac{15}{2}$
The largest integer satisfing this inequality is $7$ .
So the largest perimeter is $7 + 3cdot7 + 15 = 43 Rightarrow A$

Answer:

Problem Num : 4

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
In quadrilateral $ABCD$ , $AB = 5$ , $BC = 17$ , $CD = 5$ , $DA = 9$ , and $BD$ is an integer. What is $BD$ ?
$extbf{(A)} 11 qquad extbf{(B)} 12 qquad extbf{(C)} 13 qquad extbf{(D)} 14 qquad extbf{(E)} 15$ '

Category Triangle Inequality

Analysis

Solution/Answer
By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$ .
We got that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=oxed{13}$ .

Answer:

Problem Num : 5

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Let $T_1$ be a triangle with sides $2011, 2012,$ and $2013$ . For $n ge 1$ , if $T_n = riangle ABC$ and $D, E,$ and $F$ are the points of tangency of the incircle of $riangle ABC$ to the sides $AB, BC$ and $AC,$ respectively, then $T_{n+1}$ is a triangle with side lengths $AD, BE,$ and $CF,$ if it exists. What is the perimeter of the last triangle in the sequence $( T_n )$ ?
$extbf{(A)} frac{1509}{8} qquad extbf{(B)} frac{1509}{32} qquad extbf{(C)} frac{1509}{64} qquad extbf{(D)} fra...$
'

Category Triangle Inequality

Analysis

Solution/Answer
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.

Hence $AD=AF$ and $BD=BE$ and $CE=CF$ . Let $AD = x, BD = y$ and $CE = z$ gives three equations:
$x+y = a-1$
$x+z = a$
$y+z = a+1$
(where $a = 2012$ for the first triangle.)
Solving gives:
$x= frac{a}{2} - 1$
$y = frac{a}{2}$
$z = frac{a}{2}+1$
Subbing in gives that $T_2$ has sides of $1005, 1006, 1007$ .
$T_3$ can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with $a=1006$ ). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinately until the side lengths no longer form a triangle.
Subbing in gives $T_3$ with sides $502, 503, 504$ .
$T_4$ has sides $frac{501}{2}, frac{503}{2}, frac{505}{2}$ .
$T_5$ has sides $frac{499}{4}, frac{503}{4}, frac{507}{4}$ .
$T_6$ has sides $frac{495}{8}, frac{503}{8}, frac{511}{8}$ .
$T_7$ has sides $frac{487}{16}, frac{503}{16}, frac{519}{16}$ .
$T_8$ has sides $frac{471}{32}, frac{503}{32}, frac{535}{32}$ .
$T_9$ has sides $frac{439}{64}, frac{503}{64}, frac{567}{64}$ .
$T_{10}$ has sides $frac{375}{128}, frac{503}{128}, frac{631}{128}$ .
$T_{11}$ would have sides $frac{247}{256}, frac{503}{256}, frac{759}{256}$ but these length do not make a triangle as $frac{247}{256} + frac{503}{256} < frac{759}{256}$ .
Hence the perimeter is $frac{375}{128} + frac{503}{128} + frac{631}{128} = oxed{ extbf{(D)} frac{1509}{128}}$ $lacksquare$

Answer:

Array ( [0] => 8216 [1] => 7773 [2] => 8057 [3] => 7892 [4] => 8163 ) 5