CMC - Problem 3

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

Note by Cody Johnson
5 years, 8 months ago

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8 votes

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Is it 98?

敬全 钟 - 5 years, 8 months ago

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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 AB2+AC22AD2=2BD2=272=98A'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 AA'.

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.

Xuming Liang - 5 years, 8 months ago

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

(AB+AD)(ABAD)+(AC+AD)(ACAD)=(AB)2+(AC)22(AD)2(1)(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 AB,AD,ACA'B, A'D, A'C, we will get two new triangles, i.e. ΔABD\Delta A'BD and ΔACD\Delta A'CD. So, I apply The Theorem of Stewart,

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

Divide 7 on both sides,

(AB)2+(AC)2=2((AD)2+49)(A'B)^{2} + (A'C)^{2} = 2((A'D)^{2} + 49)

Expand RHS and subtract 2(AD)22(A'D)^{2} from both sides,

(AB)2+(AC)22(AD)2=98(A'B)^{2} + (A'C)^{2} - 2(A'D)^{2} = 98

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

(AB+AD)(ABAD)+(AC+AD)(ACAD)=98(A'B + A'D)(A'B - A'D) + (A'C + A'D)(A'C - A'D) = \boxed {98}

敬全 钟 - 5 years, 8 months ago

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Correct! 3 points for you!

Cody Johnson - 5 years, 8 months ago

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@Cody Johnson Thanks!

敬全 钟 - 5 years, 8 months ago

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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 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 ABC be a triangle such that AB=13 AB = 13 , BC=14 BC = 14 , and CA=15 CA = 15 . Let D D be the midpoint of BC \overline {BC} . Let A A' be a point coplanar to ABC \triangle ABC . Find the maximum possible value of (AB+AD)(ABAD)+(AC+AD)(ACAD). \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.

Ahaan Rungta - 5 years, 8 months ago

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

Cody Johnson - 5 years, 8 months ago

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You're welcome. Also, tags? ;)

Ahaan Rungta - 5 years, 8 months ago

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