in a quadrilateral $ABCD$, $E$ and $F$ are on $BC$ and $AD$ respectively such that the area of triangles $AED$ and $BCF$ is $\frac{4}{7}$ of the area of $ABCD$. $R$ is the intersection point of digonals of $ABCD$. $\frac{AR}{RC}=\frac{3}{5}$ and $\frac{BR}{RD}=\frac{5}{6}$. a) in what ratio does $EF$ cut the digonals?(13 points) b) find $\frac{AF}{FD}$.(5 points)
Problem
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Tags: geometry, ratio, geometric transformation, geometry proposed
oneplusone
10.08.2010 06:27
Are you sure there is nothing wrong? I think I got $E$ on the extension of $BC$.
goodar2006
10.08.2010 13:31
it was exactly our exam question and when we asked our teacher about what you have mentioned, he said that $E$ and $F$ are both on $BC$ and $AD$.
Vo Duc Dien
11.08.2011 00:51
One thing is not clear to me: "the area of triangles AED and BCF is 4/7 of the area of ABCD.... " is it the combined area of the two triangles AED and BCF = 4/7 of the area of ABCD (which is what I understand), or the area of triangle AED = area of triangle BCF = 4/7 the area of ABCD? E and F should be on the extensions of BC and AD, respectively. It does not make it clear in the problem. These problems from the non-English speaking countries really make us confuse.
SCP
17.09.2011 15:20
I suppose each area is $\frac{4}{7}$ of the area. It is easy to find that $[ABC]=\frac{5}{11}[ABCD]$ and $[BDC]=\frac{5}{8}[ABCD].$ So we calculate it is a fraction of $\frac{24}{35}$ for part $b.$ If we calculate similar for $E$ we get it isn't $\in [BC]$ so I agree with oneplusone. So for $a$ we have also a negative ratio.
Vo Duc Dien
15.12.2011 18:22
There is a flaw in the language of the problem as E and F are independent of the main configuration. One can vary by mvoing the points E and F around to make the combined area to be 4/7 of the area of ABCD to come up with a different result for the ratio AF/FD. Is there an official website for the Iranian MO to check the validity of this translation?
Vo Duc Dien
15.12.2011 18:26
Extend DA and CB to meet at N. Find the ratio of the areas of ABCD and NDC.
Vo Duc Dien
16.12.2011 20:13
This problem, initially misunderstood by some, is a beauty when all the dust is cleared away. The author should've made it clear that the area of triangle AED equals that of triangle BFC and each equals 4/7 of the area of ABCD. One must analyze carefully with numerical results to see that point F is on the interior of AD whereas point E is on the extension of BC which makes EF to cut the diagonal BD and not AC. Because of this AF/FD = 24/11.
Vo Duc Dien
16.12.2011 20:18
Let N be the intersection of the extensions of DA and CB. Find the ratio of the areas of ABCD and CDN.
Vo Duc Dien
17.12.2011 03:34
If EF cuts BD at S, the ratio is BS/SD = 11/3. This is the correct answer confirmed by software. Vo Duc Dien