Thread: X in relation to A on a triangle

1. X in relation to A on a triangle

I tried using pythagoras but I end up with equation with a lot of radicands. And to remove radicands I have to multiply each side on itself and end up with 6 defree equation.

2. This is a very confusing question.
Are you saying that all the 'red' segments are congruent?
Are you saying the base is 20 units?

3. Base is 20. A is number. X is number and yes all 3 red lines are same lenght.

4. Code:
A

x

a
D

x

C    10     B
Use right half of your triangle; change your A to a.
Extend vertical red line to base; label as shown in my diagram.
So we have: a = AB, x = AD = BD, BC = 10 (half the given 20).

So CD = sqrt(x^2 - 100)

So a^2 = [x + sqrt(x^2 - 100)]^2 + 100

A bit of simplifying will lead to: a^2 - 2x^2 = (2x)sqrt(x^2 - 100)
Square both sides:
a^4 - 4a^2 x^2 + 4x^4 = 4x^2(x^2 - 100)
Simplify:
4a^2 x^2 - 400x^2 - a^4 = 0
x^2(4a^2 - 400) = a^4

Take over...

EDIT:
above simplifies to: x = a^2 / [2sqrt(a^2 - 100]

5. Mistaken post

6. Hello, Critter314!

Base is 20. .A is a number. .X is a number.
And yes, all 3 red lines are same length.

If all this is true, the triangle is equilateral.

7. Disagree Soroban; do you mean isosceles?
Code:
D

c

d
B

a      c

C     b      A
All you need is start with any right triangle ABC; a = BC, b = AC, c = AB.

Next, extend CB to CD where BD = c; let d = AD

And cases with integer solutions exist:
as example, a = 7, b = 24, c = 25;
triangle ACD: 24-32-40, thus isosceles triangle: 40-40-48

AND c = d^2 / [2sqrt(d^2 - b^2)]
With the original problem, b = 10, AD = A and AB = x:
x = A^2 / [2sqrt(A^2 - 10^2)]

8. A bit of simplifying will lead to: a^2 - 2x^2 = (2x)sqrt(x^2 - 100)
What bit of simplifying?
I tried to simplify and I get $\displaystyle a^2-2x^2=2x\cdot(x^2-100)^2-100$

EDIT: Never mind I was wrong.