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In $ABC$, let $D$, $E$, and $F$ be points on the sides $BC$, $AC$ and $AB$, respectively, such that $BC = 4CD$, $AC=5AE$, and $AB= 6BF$.

If the area of $ABC$ is $120$, what is the area of $DEF$.

I tried connecting the vertices of the inner triangle to the opposite vertices of the outer triangle, but I think that they wont be collinear because of Ceva's Theorem, I cant think of anything else except for assuming the triangle is right and using coordinate geometry. I am asking for a nice solution

Community
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SuperMage1
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3 Answers3

3

We have $$[\triangle{CDE}]=\frac{CE}{CA}[\triangle{ADC}]=\frac{CE}{CA}\times \left(\frac{CD}{BC}[\triangle{ABC}]\right)=\frac 15[\triangle{ABC}]$$

Similarly, we get $$[\triangle{BDF}]=\frac{3}{4}[\triangle{BCF}]=\frac 34\times \frac{1}{6}[\triangle{ABC}]=\frac 18[\triangle{ABC}]$$ and $$[\triangle{AFE}]=\frac 15[\triangle{ACF}]=\frac 15\times\frac{5}{6}[\triangle{ABC}]=\frac 16[\triangle{ABC}]$$

Now note that $$[\triangle{DEF}]=[\triangle{ABC}]-[\triangle{CDE}]-[\triangle{BDF}]-[\triangle{AFE}]$$

mathlove
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2

enter image description here

well, i am a Chinese and i am worried about i can't explain it well. maybe u can look at the pic. first.

José Carlos Santos
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Ajar
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1

I use @Ajar picture and use the formula for triangle's area based on 2 sides and the angle between them.

It is easy to understand that $S_{\triangle{DEF}} = S_{\triangle{ABC}} - S_{\triangle{FBD}} - S_{\triangle{AFE}} - S_{\triangle{EDC}}(0)$

Let me assign angles $\alpha,\beta, \gamma$ for $\triangle{ABC}$

Also

$CD = a, BC = 4*a$

$AE = b, AC = 5*b$

$BF = c, AB = 6*c$

$ S_{\triangle{ABC}} = 120$

So

$S_{\triangle{AFE}}=\frac{1}{2} * AF*AE*sin\alpha = \frac{1}{2} * 5*c*b*sin\alpha(1)$

$S_{\triangle{ABC}}=\frac{1}{2} * AB*AC*sin\alpha = \frac{1}{2} * 6*c*5*b*sin\alpha(2)$

From (2) $sin\alpha=\frac{S_{\triangle{ABC}}}{15bc} (3)$

Put (3) to (1)

$S_{\triangle{AFE}}=\frac{1}{2} * 5*c*b*(\frac{S_{\triangle{ABC}}}{15bc}) = \frac{S_{\triangle{ABC}}}{6}(4)$

Do the same with $\triangle{FBD}, \triangle{EDC}$

$S_{\triangle{FBD}} = \frac{S_{\triangle{ABC}}}{2} (5)$

$S_{\triangle{EDC}} = \frac{S_{\triangle{ABC}}}{5} (6)$

From (0) $S_{\triangle{DEF}} = \frac{S_{\triangle{ABC}}}{15} = \frac{120}{15} = 8 (7)$

Questions, errors, comments, suggestions.

yW0K5o
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