A non-degenerate triangle has integer side lengths a, b, and c such that a \le b \le c. If… — GRE General Test

A non-degenerate triangle has integer side lengths $a$ , $b$ , and $c$ such that $a \le b \le c$ . If the perimeter of the triangle is $30$ and its area is an integer, what is the length of the longest side $c$ ?

Cevap: 13

Cevap

Using Heron's formula with semi-perimeter

s = 15

, the area of the triangle is

\sqrt{15(15-a)(15-b)(15-c)}

. By defining

x = 15-a

y = 15-b

, and

z = 15-c

, we have

x+y+z = 15

with

x \ge y \ge z \ge 1

. For the area to be an integer,

15xyz

must be a perfect square. The AM-GM inequality bounds

xyz \le 125

, leaving

xyz = 60

as the only valid case (

k=2

). The system

x+y+z=15

and

xyz=60

has a unique positive integer solution

(10, 3, 2)

, giving side lengths

(5, 12, 13)

. Thus, the longest side length

c

13

Adım Adım Çözüm

Find the semi-perimeter of the triangle.

s = \frac{a+b+c}{2} = \frac{30}{2} = 15

Heron's formula requires the semi-perimeter

s

to compute the area of a triangle.

Set up Heron's formula for the area.

\text{Area} = \sqrt{15(15-a)(15-b)(15-c)}

Heron's formula relates the side lengths of a triangle directly to its area.

Define variables for the differences from the semi-perimeter.

Let

x = 15-a

y = 15-b

, and

z = 15-c

. Then

x+y+z = 15

and

x \ge y \ge z \ge 1

Because

a \le b \le c

, their complements with respect to

15

satisfy

x \ge y \ge z

Determine the condition for the area to be an integer.

\text{Area} = \sqrt{15xyz}

must be an integer, so

15xyz = m^2

for some integer

m

, implying

xyz = 15k^2

for an integer

k \ge 1

For

\sqrt{15xyz}

to be an integer,

15xyz

must be a perfect square containing

15

as a factor.

Bound

xyz

using the AM-GM inequality.

xyz \le \left(\frac{x+y+z}{3}\right)^3 = 5^3 = 125 \implies 15k^2 \le 125 \implies k=1 \text{ or } k=2

The AM-GM inequality provides a maximum upper bound for the product of three positive numbers with a fixed sum.

Evaluate the possible values of

k

k=1

xyz = 15

, which has no integer solutions for

x+y+z=15

. If

k=2

xyz = 60

, which yields

(x, y, z) = (10, 3, 2)

Solving the system

x+y+z=15

and

xyz=60

under

x \ge y \ge z \ge 1

gives

z=2

y=3

x=10

uniquely.

Calculate the side lengths and identify the longest side

c

a = 5

b = 12

c = 13

, so the longest side length is

13

Since

c = 15 - z = 15 - 2 = 13

, the longest side of the right triangle

(5, 12, 13)

13

Anahtar Kavram

Heron's formula, integer partitioning, and the AM-GM inequality for triangle area optimization

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