Extension extended

Geometry Level 4

$\Delta$ ABC is an equilateral triangle with side length 1. AC, CB & AB are extended respectively to D, F & E such that E is a point on segment DF .

Given that, lengths: $\text{CD} =10 \quad,\quad \text{BF}=1.$

If the length of segment $\color{#3D99F6} {\text{BE}}$ can be expressed as $\frac ab$ where $a$ & $b$ are coprime positive integers, then find $a+b$ .

Inspiration

The answer is 11.

1 solution

Md Omur Faruque
Aug 23, 2015

Let's take a look at the very familiar parallel line dividing system of a segment.

Segment AB is divided into 2, 3 & 4 equal parts by the $\text{Black}$ , $\color{cyan} {\text{Cyan}}$ , & $\color{#D61F06} {\text {Red}}$ lines respectively.

Similarly, in this problem if we extend BE to BG where BG = 1 & draw a ray GF , then we can get the answer very easily. Here is a picture:

Notice that, $\text{AD} \parallel \text{GF}$ . As, segment AD is 11: $\boldsymbol {\color{#D61F06} {\text{EG}} =\frac{\text{AG}} {11+1}}$ $\boldsymbol {\Rightarrow \color{#D61F06} {\text{EG}} =\frac{2}{12}=\frac 16}$ Thus, $\boldsymbol {\color{#3D99F6} {\text{BE}} =\text{BG} - \color{#D61F06} {\text{EG}}}$ $\boldsymbol {\Rightarrow \color{#3D99F6} { \text{BE}} =1-\frac{1}{6}}$ $\boldsymbol {\Rightarrow \color{#3D99F6} {\text{BE}} =\frac{5}{6}}$

As, $5$ & $6$ are coprime, the answer is $\boldsymbol {5+6=}\color{#69047E} {\boxed {11}}$ .

Moderator note:

See Menelaus' Theorem for a one-line solution.

Use Menelaus' Theorem for a one-line solution.

Calvin Lin Staff - 5 years, 9 months ago

Thanks, didn't know that before.

MD Omur Faruque - 5 years, 9 months ago

Then also, you'll have to find $\dfrac{EF}{DF}$

You mean to use Menelaus' Theorem like this right?

$\dfrac{DC}{AC}\times\dfrac{BE}{AB}\times\dfrac{EF}{DF} = 1$

Kishore S. Shenoy - 5 years, 9 months ago

No, that exact approach was not how I intended to use it. They are many ways of using it, especially with the numerous number of triangles and transversals that are present in the setup. For example, taking triangle $ABC$ with transversal $EDF$ , we get that

$1 = \frac{AD}{DC} \times \frac{CF}{FB} \times \frac{BE}{EA}$

This tells us

$1 = \frac{11}{10} \times \frac{2}{1} \times \frac{ BE} { BE + 1 }$

Hence, $BE = \frac{10}{22-10}$ and we are done.

If you want to use $\frac{ED}{DF}$ (which is currently another unknown), then you will need to use additional triangle-transveral pairs like triangle CDF with transversal ABE, to eliminate it.

Calvin Lin Staff - 5 years, 9 months ago

@Calvin Lin – Oh! Yes, that's simple! Thanks!

Kishore S. Shenoy - 5 years, 9 months ago

@Calvin Lin sir, do you like Menelaus' Theorem so much?!

Kishore S. Shenoy - 5 years, 9 months ago

Extension extended

The answer is 11.

1 solution

Moderator note:

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