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In the drawing, $AM=MN=NC$ and $\frac {BP}{PC}=\frac {5}{3}$, if the area of the gray region is 8, whats the area of $\triangle ABC$?

enter image description here

I saw that $\triangle ABN$ and $\triangle BNC$ have the same height but their bases are in ratio $2:1$, and the sum of the areas of those triangles gives the final answer, but i don't know how to continue and how to use the area of the gray triangle. I think i'm close.

Any hints?

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  • $\begingroup$ Add an auxiliary segment through P parallel to AC intersecting BN at Q. Call the intersection of BN and MP R. $\frac {QP}{NC}=\frac {5}{8}$ and QPR is similar to NMR. I think that will get you half way there. $\endgroup$ – Steve B Aug 11 '18 at 2:09
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Add an auxiliary segment through P parallel to AC intersecting BN at Q and AB at S. Call the intersection of BN and MP R. Let $h_1$ be the altitude of triangle QRP
Let $h_2$ be the altitude of triangle MRN
Let $h_3$ be the altitude of triangle PMC
Let $h_4$ be the altitude of triangle BSP
Let $h_5$ be the altitude of triangle BAC

$h_3$ = $h_1$ + $h_2$
$h_5$ = $h_4$ + $h_3$

Triangle BQP is similar to triangle BNC. $\frac {BP}{BC}=\frac {5}{8}$, so $\frac {QP}{NC}=\frac {QP}{MN}=\frac {5}{8}$. Triangle QRP is similar to triangle NRM. So $h_1$ = $\frac {5}{8}$$h_2$

$h_3$ = $h_1$ + $h_2$ = $\frac {5}{8}$$h_2$ + $h_2$ = $\frac {13}{8}$$h_2$

Triangle BSP is similar to triangle BAC, so $h_4$ = $\frac {5}{8}$$h_5$

$h_5$ = $h_4$ + $h_3$ = $\frac {5}{8}$$h_5$ + $h_3$
$\frac {3}{8}$$h_5$ = $h_3$, so $h_5$ = $\frac {8}{3}$$h_3$ = $\frac {13}{3}$$h_2$

Area of RMN = 8 = $\frac {1}{2}$(MN)$h_2$

Area of BAC = $\frac {1}{2}$(3MN)$h_5$ = $\frac {1}{2}$(3MN)$\frac {13}{3}$$h_2$ = 13($\frac {1}{2}$(MN)$h_2$) = (13)(8) = 104

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Divide BC by 16 segment with same size with paralell line to PM. Then, draw 4 lines paralell to BN such that one line contains A, one line contains M, one line contains N and one line contains C. You give 48 parallelogram with same area. Each parallelogram area has $\frac{16}{3}$ The $\Delta ABC$ area is $39\cdot\frac{16}{6} = 104$

Click here, Solution Image

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First of all, consider that not all triangles that meet the constraints are similar to each other. The angle $\angle C$ is can be almost anything, as can the ratio of $AC$ to $BC$.

What this means is that you can choose a specific triangle that makes calculating the total area easy. There are various choices, but once you have concrete values for these, it is relatively easy to calculate the area of the entire triangle. For example choose them such that $AC = BC$ and $\angle MNB$ is a right angle.

Then all you need to do is prove that all of these triangles have the same area. If you just do the calculations with a few different values, you can convince yourself of this, or you can use properties of affine transformations. In general, if the ratios of lengths are preserved, then the ratios of areas are preserved as well.

Hope that helps.

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