# 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
4 years, 1 month ago

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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.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

Nice use of Barycentric Coordinates !

- 4 years, 1 month ago

@Xuming Liang @Calvin Lin @Sualeh Asif Here are some simple problems ! Presently not very good at Geometry, but planning to improve.

- 4 years, 1 month ago