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In a triangle $\triangle ABC$ the medians $AM$ And $CN$ to the sides $BC$ and $AB$ respectively,

intersects at the point $O$.

Let $P$ be the the midpoint of the side $AC$ and let $MP$ intersects $CN$ at $Q$.

If the area of the triangle $\triangle OMQ$ is a 's' square units, then the area of the triangle $\triangle ABC$ is

(a)16s (b) 18s (c) 21s (d)24s enter image description here

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  • $\begingroup$ It is allowed to use a ruler for drawing straight lines. $\endgroup$ – Jean Marie Apr 9 '17 at 7:07
  • $\begingroup$ @JeanMarie Was that a sarcastic comment? $\endgroup$ – Babai Apr 9 '17 at 14:06
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    $\begingroup$ @Babai: you are not a novice anymore, Babai. A sound geometry question should include your attempts and possibly a decent diagram; Geogebra is fairly simple to use. $\endgroup$ – Jack D'Aurizio Apr 9 '17 at 16:02
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There is a short proof using barycentric coordinates (see formula 4 in (http://mathworld.wolfram.com/BarycentricCoordinates.html)):

$$\dfrac{area(OMQ)}{area(ABC)}=\begin{vmatrix}\tfrac{1}{3}&0&\tfrac{1}{4}\\\tfrac{1}{3}&\tfrac{1}{2}&\tfrac{1}{4}\\\tfrac{1}{3}&\tfrac{1}{2}&\tfrac{1}{2}\end{vmatrix}=\frac{1}{24}$$

where the columns of the determinant are the barycentric coordinates of O,M,Q, resp. For example the barycentric coordinates of $O$ are $\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3}$ because $O$ can be expressed as $O=\tfrac{1}{3}A+\tfrac{1}{3}B+\tfrac{1}{3}C$. The same for the other two points.

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Let $CO\cap AB=\{K\}$ and $KO=2x$.

Hence, $OC=4x$, $KC=6x$, $QC=3x$, which gives $OQ=x$.

Now, since $OQ=\frac{1}{4}OC$, we obtain $S_{\Delta MOC}=4$

and since $OM=\frac{1}{3}AM$, we get $S_{\Delta AMC}=12$ and

since $BM=MC$, we obtain $S_{\Delta ABC}=24$.

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  • $\begingroup$ CO and BC intersect only at C. What is your K? $\endgroup$ – Babai Apr 9 '17 at 6:13
  • $\begingroup$ @Babai It was typo. See now. $\endgroup$ – Michael Rozenberg Apr 9 '17 at 6:19
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    $\begingroup$ Why is $QC=3x$? $\endgroup$ – Babai Apr 9 '17 at 6:29
  • $\begingroup$ @Babai Because $MP$ is a midline of $\Delta ABC$, which gives $KQ=QC$ and since $KC=6x$, we obtain $QC=3x$. $\endgroup$ – Michael Rozenberg Apr 9 '17 at 6:43
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    $\begingroup$ Wouldn't $OQ=\frac14 OC$? $\endgroup$ – browngreen Apr 9 '17 at 6:57
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Yet another solution: $$S_{ABC} = 2 S_{AMC} = 4 S_{AMP} = 12 S_{OMP} = 24 S_{OMQ}.$$ This is because $BC = 2MC$, $AC= 2AP$, $AM=3OM$, and $MP=2MQ$ respectively.

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