**Problem 3. (3 points)** In \(\triangle ABC\), let \(D\) be the midpoint of \(\overline{BC}\). If \(AB=13\), \(BC=14\), and \(CA=15\), find the maximum value of \((A'B+A'D)(A'B-A'D)+(A'C+A'D)(A'C-A'D)\), where \(A'\) can be any point in the plane of \(\triangle ABC\).

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TopNewestIs it 98?

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Yes, by the Median length theorem aka Apollonius' Theorem(you can call it a special case of Stewarts but the proof is trivial using cosine rule), the expression is equal to \(A'B^2+A'C^2-2A'D^2=2BD^2=2*7^2=98\). This means that the expression stays invariant regardless of the location of \(A'\).

More on the theorem: http://en.wikipedia.org/wiki/Apollonius'_theorem

Sometimes being well-known to these theorems and the structure of some equations can really help.

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I do in this way. First, I expand the given expression,

\((A'B + A'D)(A'B - A'D) + (A'C + A'D)(A'C - A'D) = (A'B)^{2} + (A'C)^{2} - 2(A'D)^{2} - (1)\)

When we connect \(A'B, A'D, A'C\), we will get two new triangles, i.e. \(\Delta A'BD\) and \(\Delta A'CD\). So, I apply The Theorem of Stewart,

\(7(A'B)^{2} + 7(A'C)^{2} = 14((A'D)^{2} +49)\)

Divide 7 on both sides,

\((A'B)^{2} + (A'C)^{2} = 2((A'D)^{2} + 49)\)

Expand RHS and subtract \(2(A'D)^{2}\) from both sides,

\((A'B)^{2} + (A'C)^{2} - 2(A'D)^{2} = 98\)

Comparing the equation above with \((1)\), we get

\((A'B + A'D)(A'B - A'D) + (A'C + A'D)(A'C - A'D) = \boxed {98}\)

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Correct! 3 points for you!

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Too sleepy to solve it right now, but a little piece of humble advice on the problem-writing. The way you have defined \( A' \) is not logically incorrect but it is a bit confusing and it is not the standard way of writing it. Here is how I would rephrase the problem. Also, "in" is not completely rigorous.

Let \( ABC \) be a triangle such that \( AB = 13 \), \( BC = 14 \), and \( CA = 15 \). Let \( D \) be the midpoint of \( \overline {BC} \). Let \( A' \) be a point coplanar to \( \triangle ABC \). Find the maximum possible value of \[ \left( A'B + A'D \right) \cdot \left(A'B - A'D\right) + \left(A'C + A'D\right) \cdot \left(A'C - A'D\right). \]

Hope you don't take this as criticism or egotistical advice; rather, as suggestions on problem-writing.

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I had that at first but was tentative. Thanks for your criticism, I genuinely appreciate it.

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You're welcome. Also, tags? ;)

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