In triangle ABC, medians AM and BK are 9 and 12, side AB = 10. What is the third median?

Triangle ABC, AM = 9 – median BC, BK = 12 – median AC, CE – median AB, AB = 10. The length of the median of the triangle is calculated by the formula:
m = √ (2b ^ 2 + 2c ^ 2 – a ^ 2) / 2,
where a is the side to which the median is drawn, b and c are the other sides of the triangle.
1. Let us express all three medians through the formulas:
AM = √ (2AB ^ 2 + 2AC ^ 2 – BC ^ 2) / 2;
ВK = √ (2AB ^ 2 + 2BC ^ 2 – AC ^ 2) / 2;
CE = √ (2BC ^ 2 + 2AC ^ 2 – AB ^ 2) / 2.
Let’s substitute the known data:
9 = √ (2 * 10 ^ 2 + 2АС ^ 2 – ВС ^ 2) / 2;
12 = √ (2 * 10 ^ 2 + 2BC ^ 2 – AC ^ 2) / 2;
CE = √ (2BC ^ 2 + 2AC ^ 2 – 10 ^ 2) / 2.
2. Let AC = x, BC = y, then we get a system of equations with three unknowns:
√ (200 + 2x ^ 2 – y ^ 2) / 2 = 9;
√ (200 + 2y ^ 2 – x ^ 2) / 2 = 12;
√ (2y ^ 2 + 2x ^ 2 – 100) / 2 = CE.
3. In the first equation of the system, we express y ^ 2 through x ^ 2:
√ (200 + 2x ^ 2 – y ^ 2) = 18 (proportional);
200 + 2x ^ 2 – y ^ 2 = 324;
– y ^ 2 = 324 – 200 – 2x ^ 2;
y ^ 2 = 2x ^ 2 – 124.
4. Substitute the resulting expression into the second and third equations of the system:
√ (200 + 2 (2x ^ 2 – 124) – x ^ 2) / 2 = 12;
√ (2 (2x ^ 2 – 124) + 2x ^ 2 – 100) / 2 = CE.
We got a system of equations with two unknowns:
√ (200 + 4x ^ 2 – 248 – x ^ 2) / 2 = 12;
√ (4x ^ 2 – 248 + 2x ^ 2 – 100) / 2 = CE.
Here are similar ones:
√ (3x ^ 2 – 48) / 2 = 12;
√ (6x ^ 2 – 348) / 2 = CE.
5. In the first equation of the system, we find the value of x:
√ (3x ^ 2 – 48) = 24 (proportional);
3x ^ 2 – 48 = 576;
3x ^ 2 = 576 + 48;
3x ^ 2 = 624;
x ^ 2 = 624/3;
x ^ 2 = 208.
6. Substitute the resulting value into the second equation of the system:
√ (6 * 208 – 348) / 2 = CE.
Let’s solve the equation with one unknown:
CE = √ (1248 – 348) / 2;
CE = √ (900) / 2;
CE = 30/2;
CE = 15.
Answer: The third median of the CE is 15.