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In the figure, CD=2AB=2BC and FE = ED

Find AG: HE

This is an Olympic question in China, I tried, still can't figure it out. Please.

esoteric elf
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  • can you see that $C$ is the midpoint of $AD$, and $B$ is the midpoint of $AC$? Also, $E$ is the midpoint of $FD$ – Varun Iyer Aug 06 '14 at 19:12

4 Answers4

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Hint: Note that EC is half FA and parallel to it. So HE:HA and FH:FC ratios should be clear. Then apply Menelaus' theorem for triangle AHC slashed by line FGB in order to infer ratio AG:GH.

ir7
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  • I got to the ratios $HE:HA$ and $FH:FC$ by noting that both $AE$ and $FC$ are medians. – Hao Ye Aug 06 '14 at 21:10
  • @HaoYe Good. Menelaus then says: FC:FH X HG:GA X AB:BC = 1. (Remember that you already know AB:BC.) So now, you know both HE:HA and HG:GA. The answer follows then easily. – ir7 Aug 06 '14 at 21:35
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    Yeah, I was going to suggest using mass points on $\triangle AFC$, but I think it boils down to Menelaus' anyway. – Hao Ye Aug 06 '14 at 22:09
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since the point $F$ is free to move, then we may choose the triangle $ADF$ to be right-angled and isosceles. in which case $AG=HE$. however this does not prove the general result unless we can show this ratio invariant under translations of $F$

David Holden
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Since $2|AB| = 2|BC| = |CD|$ it follows that $C$ is the midpoint of $AD$. By definition, we also know that $E$ is the midpoint of $FD$. It follows that $FC$ and $AE$ are medians of triangle $\triangle FAD$ and hence $H$ is the centroid of the triangle. The centroid divides any median into a ratio of $1:2$, hence $|AH| = 2|HE|$.

Now notice that the points $A,G,H,E$ and $A,B,C,D$ are related by a perspectivity centered at $F$. It follows that the cross-ratio is invariant between the two point sets.

Therefore we have: $$3=\frac{|AC||DB|}{|DC||AB|} = (A,D; C,B) = (A,E; H,G) = \frac{|AH||EG|}{|EH||AG|}=2\frac{|EG|}{|AG|}$$ $$\frac{1}{2}=\frac{|AB||CD|}{|CB||AD|}=(A,C;B,D)=(A,H; G,E)=\frac{|AG||HE|}{|HG||AE|} = \frac{1}{3}\frac{|AG|}{|HG|}$$ Hence we have $$\frac{|EG|}{|AG|} = \frac{|AG|}{|HG|}= \frac{3}{2}$$ Finally, we have $$\frac{3}{2}=\frac{|EG|}{|AG|} = \frac{|GH|}{|AG|}+\frac{|HE|}{|AG|} = \frac{2}{3} + \frac{|HE|}{|AG|}$$ Therefore $$\frac{|AG|}{|HE|} = \frac{6}{5}$$

EuYu
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It is possible to approach the problem through Ceva's and Van Obel's theorems.

Call $G',H'$ the points in which $DG,DH$ cut $AF$. Due to Ceva's theorem we have: $$\frac{AG'}{G'F}=\frac{1}{3},\qquad \frac{AH'}{H'F}=1,$$ hence Van Obel's theorem gives: $$\frac{AG}{GE}=\frac{2}{3},\qquad \frac{AH}{HE}=2,$$ and we have: $$ AG:GH:HE = 6:4:5,$$ hence $\frac{AG}{HE}=\frac{6}{5}$.

Jack D'Aurizio
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    this is surprisingly easier than i thought, i tried work around with area sum and difference but i guessed i messed up, thanks so much – esoteric elf Aug 08 '14 at 17:14