2022 SSMO Speed Round Problems/Problem 3: Difference between revisions
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==Solution== | ==Solution== | ||
We have | |||
<cmath>\begin{align*} | |||
[AEF] &= [AEC]-[CEF]\\ | |||
&= \frac{CE}{CD}\cdot[ACD]-\frac{CE}{CD}\cdot\frac{CE}{CE+AB}\cdot[ACD]\\ | |||
&= \frac{CE}{CE+DE}\cdot[ACD]-\frac{CE}{CE+DE}\cdot\frac{CE}{CE+CD}\cdot[ACD]\\ | |||
&= \frac{2}{2+3}\cdot[ACD]-\frac{2}{2+3}\cdot\frac{CE}{CE+(CE+DE)}\cdot[ACD]\\ | |||
&= \frac{2}{5}\cdot[ACD]-\frac{2}{5}\cdot\frac{2}{2+(2+3)}\cdot[ACD]\\ | |||
&= \left(\frac{2}{5}-\frac{2}{5}\cdot\frac{2}{7}\right)[ACD]\\ | |||
&= \frac{2}{7}[ACD] = \frac{2}{7}\cdot\frac{1}{2}\cdot[ABCD] = \frac{1}{7}[ABCD]\implies\\ | |||
[ABCD] &= 7[AEF] = 7\cdot15 = \boxed{105}. | |||
\end{align*}</cmath> | |||
~pinkpig | |||
Latest revision as of 16:18, 13 September 2025
Problem
Let 
be a parallelogram such that 
is a point on 
such that 
Suppose that 
and 
intersect at 
If the area of triangle 
is 
find the area of .
Solution
We have
![\begin{align*} [AEF] &= [AEC]-[CEF]\\ &= \frac{CE}{CD}\cdot[ACD]-\frac{CE}{CD}\cdot\frac{CE}{CE+AB}\cdot[ACD]\\ &= \frac{CE}{CE+DE}\cdot[ACD]-\frac{CE}{CE+DE}\cdot\frac{CE}{CE+CD}\cdot[ACD]\\ &= \frac{2}{2+3}\cdot[ACD]-\frac{2}{2+3}\cdot\frac{CE}{CE+(CE+DE)}\cdot[ACD]\\ &= \frac{2}{5}\cdot[ACD]-\frac{2}{5}\cdot\frac{2}{2+(2+3)}\cdot[ACD]\\ &= \left(\frac{2}{5}-\frac{2}{5}\cdot\frac{2}{7}\right)[ACD]\\ &= \frac{2}{7}[ACD] = \frac{2}{7}\cdot\frac{1}{2}\cdot[ABCD] = \frac{1}{7}[ABCD]\implies\\ [ABCD] &= 7[AEF] = 7\cdot15 = \boxed{105}. \end{align*}](http://latex.artofproblemsolving.com/d/b/f/dbf1e084e742b8f3777b189b71b8beb0dd732702.png)
~pinkpig
