In \triangle ABC, the lengths of sides AB and AC are both 13. A point D lies on side BC su… — ACT

In $\triangle ABC$ , the lengths of sides $AB$ and $AC$ are both $13$ . A point $D$ lies on side $BC$ such that $AD$ is an integer. If the perimeter of $\triangle ABD$ is equal to the perimeter of $\triangle ACD$ , what is the sum of all possible integer values for the length of $BC$ ?

Cevap: 34

Cevap

The perimeters of

\triangle ABD

and

\triangle ACD

are equal, and since

AB = AC = 13

, this forces

BD = CD

, making

D

the midpoint of

BC

. In the isosceles triangle

\triangle ABC

, the median

AD

is perpendicular to

BC

, making

\triangle ABD

a right triangle. By the Pythagorean theorem,

BD^2 + AD^2 = 169

. Since

AD

is an integer,

BD

must also be an integer (a half-integer would result in

AD^2

ending in

.25

, which cannot be a perfect square of an integer). The only positive integer solutions for

(BD, AD)

are

(5, 12)

and

(12, 5)

. This results in

BC = 2 \cdot BD

being either

10

24

. The sum of these possible values is

10 + 24 = 34

Adım Adım Çözüm

Set the perimeters of

\triangle ABD

and

\triangle ACD

equal to each other.

BD = CD

Since

AB = AC = 13

, equating

AB + BD + AD = AC + CD + AD

simplifies directly to

BD = CD

Determine the relationship between

AD

and

BC

\triangle ABD

is a right triangle with hypotenuse

13

In an isosceles triangle, the median to the base is also the altitude, so

AD \perp BC

Apply the Pythagorean theorem to

\triangle ABD

BD^2 + AD^2 = 169

The sum of the squares of the legs in right triangle

\triangle ABD

must equal the square of the hypotenuse

AB = 13

Analyze the parity and integer constraints of

BD

and

AD

BD

must be a positive integer.

BD

were a half-integer,

BD^2

would end in

.25

, preventing

AD^2

from being an integer, which contradicts the given condition that

AD

is an integer.

Identify the Pythagorean triples with a hypotenuse of

13

(BD, AD) \in \{(5, 12), (12, 5)\}

The only positive integer solutions to

x^2 + y^2 = 13^2

are

(5, 12)

and

(12, 5)

Calculate the possible lengths of

BC

and sum them.

BC \in \{10, 24\}

, and their sum is

34

Since

D

is the midpoint of

BC

, the length of

BC

2 \cdot BD

, yielding

2 \cdot 5 = 10

and

2 \cdot 12 = 24

. Both satisfy the triangle inequality because

BC < AB + AC = 26

Anahtar Kavram

Properties of Isosceles Triangles and the Pythagorean Theorem

Tahmini Süre:3m 0s

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