help!!

Problem 1

 

 

Since the perimeter of the triangle is 50, the length of the base must be even, since the sum of the length of the base and one of the legs must be less than 50. If the length of the base is 2x, then the length of the legs must be 25-x, since the perimeter of the triangle is 50. Since 25-x must be greater than x, x must be less than 12.5. Let's check the values of x. If x=1, then the length of the legs is 24, which is even. The triangle is equilateral (all sides are equal) and therefore there is only one non-congruent triangle. If x=2, then the length of the legs is 23, which is odd. The triangle is scalene (all sides are different) and therefore there are 3 non-congruent triangles. If x=3, then the length of the legs is 22, which is even. The triangle is isosceles (two sides are equal) and therefore there are 2 non-congruent triangles. If x=4, then the length of the legs is 21, which is odd. The triangle is scalene (all sides are different) and therefore there are 3 non-congruent triangles. If x=5, then the length of the legs is 20, which is even. The triangle is isosceles (two sides are equal) and therefore there are 2 non-congruent triangles. If x=6, then the length of the legs is 19, which is odd. The triangle is scalene (all sides are different) and therefore there are 3 non-congruent triangles. If x=7, then the length of the legs is 18, which is even. The triangle is right (two sides are equal and the third side is the hypotenuse) and therefore there are 2 non-congruent triangles. If x=8, then the length of the legs is 17, which is odd. The triangle is scalene (all sides are different) and therefore there are 3 non-congruent triangles. If x=9, then the length of the legs is 16, which is even. The triangle is right (two sides are equal and the third side is the hypotenuse) and therefore there are 2 non-congruent triangles. If x=10, then the length of the legs is 15, which is odd. The triangle is isosceles (two sides are equal) and therefore there are 2 non-congruent triangles. If x=11, then the length of the legs is 14, which is even. The triangle is isosceles (two sides are equal) and therefore there are 2 non-congruent triangles. If x=12, then the length of the legs is 13, which is odd. The triangle is scalene (all sides are different) and therefore there are 3 non-congruent triangles. Therefore, there are a total of 20 non-congruent triangles with integer side lengths and perimeter 50.

 

Problem 2

 

There are 3 ways to choose one number from the set {1, 2, 3}.

There are 3 ways to choose one number from the set {3, 4, 5}.

There are 3 ways to choose one number from the set {5, 6, 7}.

However, not all combinations of these three numbers will form a triangle. In order for three numbers to form a triangle, the sum of any two sides must be greater than the third side.

So, we need to check each combination to see if it forms a triangle.

1, 3, 5: Yes 1, 3, 6: No 1, 3, 7: No 1, 4, 5: Yes 1, 4, 6: No 1, 4, 7: No 1, 5, 6: Yes 1, 5, 7: Yes 2, 3, 5: Yes 2, 3, 6: No 2, 3, 7: No 2, 4, 5: Yes 2, 4, 6: Yes 2, 4, 7: No 2, 5, 6: Yes 2, 5, 7: Yes 3, 4, 5: Yes 3, 4, 6: Yes 3, 4, 7: No 3, 5, 6: Yes 3, 5, 7: Yes 3, 6, 7: No

So, there are 16 ways to choose one number from each of the three sets such that the three numbers could be the sides of a nondegenerate triangle.

 

Problem 3

 

To find the longest possible integer length of the third altitude, we can use the following steps:

Let the length of the third altitude be h.

Let the lengths of the sides opposite the altitudes of length 12 and 15 be a and b, respectively.

We know that the area of a triangle is equal to half the product of its base and height. Therefore, we can write the following equations:

Area of the triangle = $\frac{1}{2}ab = \frac{1}{2} \cdot 12h = \frac{1}{2} \cdot 15h

Solving for h, we get:

h = \frac{2ab}{12 + 15} = \frac{2ab}{27}

Since a, b, and h are all integers, the longest possible integer length of h is equal to the greatest common divisor (GCD) of 2ab and 27.

To find the GCD of 2ab and 27, we can use the Euclidean algorithm:

GCD(2ab, 27) = GCD(2ab - 27a, 27) = GCD(27a - 6ab, 27) = ... = GCD(27, 27) = 27

Therefore, the longest possible integer length of the third altitude is 27.