Synthetic Geometry Group - Karthik's Proposal

Here are my submissions :

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

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

Note by Karthik Venkata
3 years, 11 months ago

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

Notice that FF has mass p+q+rp+q+r. This implies that MFMB=qp+q+r\frac{MF}{MB} = \frac{q}{p+q+r} and we are done.

Alan Yan - 3 years, 10 months ago

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Nice use of Barycentric Coordinates !

Karthik Venkata - 3 years, 10 months ago

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

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

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

Alan Yan - 3 years, 10 months ago

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

Karthik Venkata - 3 years, 10 months ago

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

Karthik Venkata - 3 years, 11 months ago

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