If a2 + b2 + c2 = ab + bc + ac, then the value of \(\rm \frac{11 a^4+13 b^4+17 c^4}{17 a^2 b^2+9 b^2 c^2+15 c^2 a^2}\) is ?
1
Understanding the Given Algebraic Equation
The problem provides an algebraic equation and asks for the value of a specific expression based on this condition. The given condition is:
\(a^2 + b^2 + c^2 = ab + bc + ac\)
We need to determine the implications of this equation on the relationship between \(a\), \(b\), and \(c\) to evaluate the expression:
\(\frac{11 a^4+13 b^4+17 c^4}{17 a^2 b^2+9 b^2 c^2+15 c^2 a^2}\)
Deriving the Relationship between a, b, and c
Let's rearrange the given equation to find the relationship between \(a\), \(b\), and \(c\). The equation is:
\(a^2 + b^2 + c^2 - ab - bc - ac = 0\)
We can multiply the entire equation by 2 to transform it into a sum of squares. Multiplying by 2 gives:
\(2(a^2 + b^2 + c^2 - ab - bc - ac) = 2(0)\)
\(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0\)
Now, let's regroup the terms:
\((a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) = 0\)
Recognize that each group of terms is a perfect square trinomial:
- \(a^2 - 2ab + b^2 = (a-b)^2\)
- \(b^2 - 2bc + c^2 = (b-c)^2\)
- \(c^2 - 2ac + a^2 = (c-a)^2\)
Substituting these back into the equation:
\((a-b)^2 + (b-c)^2 + (c-a)^2 = 0\)
For real numbers \(a\), \(b\), and \(c\), the square of a real number is always non-negative (greater than or equal to zero). The sum of non-negative numbers can only be zero if each individual number is zero.
Therefore, we must have:
- \((a-b)^2 = 0 \implies a-b = 0 \implies a = b\)
- \((b-c)^2 = 0 \implies b-c = 0 \implies b = c\)
- \((c-a)^2 = 0 \implies c-a = 0 \implies c = a\)
Combining these results, we conclude that \(a = b = c\).
Evaluating the Algebraic Expression
Now that we know \(a = b = c\), we can substitute this relationship into the given expression:
\(\frac{11 a^4+13 b^4+17 c^4}{17 a^2 b^2+9 b^2 c^2+15 c^2 a^2}\)
Let's substitute \(b=a\) and \(c=a\) into the numerator and the denominator.
Numerator:
\(11 a^4 + 13 b^4 + 17 c^4\)
Substitute \(b=a\) and \(c=a\):
\(11 a^4 + 13 a^4 + 17 a^4\)
Combine the terms:
\((11 + 13 + 17) a^4 = 41 a^4\)
Denominator:
\(17 a^2 b^2 + 9 b^2 c^2 + 15 c^2 a^2\)
Substitute \(b=a\) and \(c=a\):
\(17 a^2 (a^2) + 9 (a^2) (a^2) + 15 (a^2) a^2\)
\(17 a^4 + 9 a^4 + 15 a^4\)
Combine the terms:
\((17 + 9 + 15) a^4 = 41 a^4\)
Now, substitute the simplified numerator and denominator back into the expression:
\(\frac{41 a^4}{41 a^4}\)
Assuming \(a \neq 0\), we can cancel out \(41 a^4\):
\(\frac{41 a^4}{41 a^4} = 1\)
If \(a=0\), then \(b=c=0\) as well. In this case, the expression becomes \(\frac{0}{0}\), which is an indeterminate form. However, in the context of such problems with multiple-choice numerical options, it is implied that we are looking for a unique value that the expression takes under the given condition for non-trivial cases (where the denominator is not zero). The derivation \(a=b=c\) holds, and for any non-zero value of \(a\) (and thus \(b\) and \(c\)), the value is 1.
Conclusion
Given the condition \(a^2 + b^2 + c^2 = ab + bc + ac\), we deduced that \(a = b = c\) (for real numbers). Substituting this into the given expression, the numerator becomes \(41a^4\) and the denominator becomes \(41a^4\). Therefore, the value of the expression is 1.
| Step | Description | Result |
|---|---|---|
| 1 | Start with the given condition | \(a^2 + b^2 + c^2 = ab + bc + ac\) |
| 2 | Rearrange and multiply by 2 | \(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0\) |
| 3 | Group terms into perfect squares | \((a-b)^2 + (b-c)^2 + (c-a)^2 = 0\) |
| 4 | Deduce relationship (for real numbers) | \(a = b = c\) |
| 5 | Substitute into numerator expression | \(11 a^4 + 13 a^4 + 17 a^4 = 41 a^4\) |
| 6 | Substitute into denominator expression | \(17 a^4 + 9 a^4 + 15 a^4 = 41 a^4\) |
| 7 | Evaluate the full expression | \(\frac{41 a^4}{41 a^4} = 1\) (assuming \(a \neq 0\)) |
Revision Table: Key Concepts
| Concept | Explanation | Relevance to Problem |
|---|---|---|
| Algebraic Identity | An equation that is true for all values of the variables involved. | The condition \(a^2 + b^2 + c^2 - ab - bc - ac = 0\) is related to a standard identity. |
| Sum of Squares | The sum of terms squared, like \((x-y)^2 + (y-z)^2\). | For real numbers, a sum of squares equals zero only if each term being squared is zero. |
| Perfect Square Trinomial | An algebraic expression of the form \(x^2 \pm 2xy + y^2\), which factors as \((x \pm y)^2\). | Used to transform the given condition into a sum of squares. |
Additional Information: Generalizing the Identity
The identity \(a^2 + b^2 + c^2 = ab + bc + ac\) implying \(a=b=c\) is a specific case. A more general form related to this is:
\(a^2 + b^2 + c^2 - ab - bc - ac = \frac{1}{2} [(a-b)^2 + (b-c)^2 + (c-a)^2]\)
This identity shows the equivalence between the left side being zero and the sum of squares being zero. This is fundamental in various areas of mathematics, including inequalities and geometry.
For instance, it relates to the triangle inequality or properties of equilateral triangles when considering geometric interpretations involving distances.
Understanding this identity allows for quickly determining that the variables must be equal when the sum of squares equals zero, significantly simplifying complex expressions that depend on this condition.
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