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# Synthetic Geometry Group - Karthik's Proposal

Here are my submissions :

1) Let $$\Delta ABC$$ have cevians $$\overline{AD}$$ and $$\overline{CE}$$, which meet at a point $$F$$ inside the triangle. Prove that $$[\Delta ABC] \cdot [\Delta DEF] = [\Delta BDE] \cdot [\Delta AFC]$$, where $$[ A ]$$ denotes area of figure $$A$$.

2) If a cevian $$\overline{AQ}$$ of an equilateral triangle $$ABC$$ is extended to meet the circumcircle at $$P$$, prove that $$\dfrac{1}{\overline{PB}} + \dfrac{1}{\overline{PC}} = \dfrac{1}{\overline{PQ}}$$.

Note by Karthik Venkata
1 year, 10 months ago

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Question 1: Extend segment BF to hit AC at T. Construct $$DE$$ and call its intersection point with $$BE$$ , $$M$$. Notice that the desired relation is equivalent to $$\frac{[DEF]}{[BDE]} = \frac{[AFC]}{[ABC]} \implies \frac{FM}{MB} = \frac{TF}{TB}$$. We will use mass points. Put weights of $$p, q, r$$ on $$A, B, C$$ respectively. This implies that $$T$$ has a mass of $$p+r$$ with implies that $\frac{TF}{BF} = \frac{q}{p+r} \implies \frac{TF}{TB} = \frac{q}{p+q+r}$

Notice that $$F$$ has mass $$p+q+r$$. This implies that $\frac{MF}{MB} = \frac{q}{p+q+r}$ and we are done. · 1 year, 10 months ago

Nice use of Barycentric Coordinates ! · 1 year, 10 months ago

Question 2: We know by Ptomely's, that $$AP = PB + PC$$.

Then we just need to prove that $\frac{1}{PB} + \frac{1}{PC} = \frac{AP}{PB \cdot PC} = \frac{1}{PQ} \implies \frac{BP}{AP} = \frac{PQ}{CP}$

Observe that $$\angle ABP = \angle ACB = 60^{\circ}$$ and $$\angle APC = \angle ABC = 60^{\circ}$$ because they substend the same arc. Similarly, $$\angle CBP = \angle CAP$$ substend the same arc. This implies that $$\triangle BPQ \sim APC$$ and the desired ratio is given with these proportional lengths. · 1 year, 10 months ago

@Alan Yan Do check problem no. 5 in RMO board-2 by Nihar Mahajan ! It's more interesting than these :). Link · 1 year, 10 months ago