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I would like some help to solve this:

Consider a triangle $\triangle ABC$ with $\angle A$ a right angle and $BC=20$. Divide $BC$ into four congruent segments, that is, take the points $P,Q,R\in BC$ such that $BP=PQ=QR=RC=5$. Then, compute $AP^2+AQ^2+AR^2$.

Thanks for a while.

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Let $\theta_1=\angle APB$. By the Cosine Law, $$AB^2=AP^2+25-2(AP)(5)\cos\theta_1.$$ Let $\theta_2=\angle APQ$. Note that $\cos\theta_2=-\cos\theta_1$, since $\theta_1+\theta_2=180^\circ$. So by the Cosine Law, $$AQ^2=AP^2+25+2(AP)(5)\cos\theta_1.$$

Add up. We get $$AB^2+AQ^2=2AP^2+50.\tag{$1$}$$ Note the pretty cancellation!

We played a certain game at $P$. Play the same game at $R$. We get $$AC^2+AQ^2=2AR^2+50.\tag{$2$}$$ Add $(1)$ and $(2)$, and use the Pythagorean Theorem. We get $400+2AQ^2=2AP^2+2AR^2+100$, and dividing by $2$, $$AP^2+AR^2=AQ^2+150.$$

Play a similar game at $Q$, but using the double triangles $AQB$ and $AQC$. We get $$AB^2+AC^2=2AQ^2+200,$$ so $AQ^2=100$. (This calculation is more standard: it is how we obtain a formula for the length of a median, given the sides of a triangle.)

Thus $AP^2+AR^2=250$, and therefore $AP^2+AQ^2+AR^2=350$.

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Nice! I will check if I have the Cosine Law at this moment. –  Sigur Sep 15 '12 at 13:50
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Hold the line $BC$ fixed and allow $A$ to move; it describes a circle of radius $10$ and centre $Q$, with $BC$ as a diameter. Extend $AQ$ to meet the other side of the circle at $S$; then $PARS$ is a parallelogram with centre at $Q$ whose diagonals have lengths $10$ (for $PR$) and $20$ (for $AS$). By the parallelogram law we have $|AP|^2+|AR|^2=\frac12(10^2+20^2)=250$, and $|AQ|=10$, so the desired sum is $350$.

Added: Here’s a diagram that may help:

enter image description here

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Thanks. The parallelogram law can be proved without cosine law, right? I have to decide what solution is more elementary than the other one. –  Sigur Sep 15 '12 at 13:53
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@Sigur: It can be proved easily using coordinates; synthetic proofs are possible but a good deal harder. –  Brian M. Scott Sep 15 '12 at 18:59
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