Geometry - length of triangle question

Question

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Could anyone please help me with this question

I can't seem to solve for the ?.

The answer provided for ? is 1.

Thanks!

Answer 1

Let us guess $\Delta ABC$ is an equilateral triangle with side length of 6. F and D are the mid points of AC and BC respectively, CD=DB=AF=FC=3. Find a point E between D and B with DE=1.

If we can prove EF,AD and CH intercept at the same point G, then DE=1 is the solution.

Set some coordinates:

A(0,0),B(6,0),C(3,3$\sqrt3$), D(4.5,1.5$\sqrt3$), E(5,$\sqrt3$),F(1.5,1.5$\sqrt3$)

AE: y=$\frac{\sqrt3}{5}x$ (I)

AD:$\frac{\sqrt3}{3}x$ (II)

FB: $-\sqrt3/3x$+2$\sqrt3$ (III)

EF: $-\sqrt3/7x$+12$\sqrt3/7$ (IV)

Solve (I) and (III) to get the intercept point of AE and FB at H($\frac{15}{4},\frac{3\sqrt3}{4}$), then

HC: $-3\sqrt3x+12\sqrt3$ (V)

Solve (II) and (V) to get the intercept point of AD and HC at G$_1$(3.6,$\frac{6\sqrt3}{5}$)

Solve (II) and (IV) to get the intercept point of AD and EF intercept at G$_2$(3.6,$\frac{6\sqrt3}{5}$)

We proved AD,EF and HC intercept at the same point $G_1=G_2=G$.

Now one can conclude the answer is indeed DE=1.

Answer 2

There is not enough information to solve this problem, because you gave us just one side of the triangle without any angle, this leaves us with nothing to solve

But for the sake of a solution, let's assume the extra information is hidden in the triangle, maybe similarities, angles, length, etc.......... But when I checked, no lines were parallel, no smaller triangles were similar, some angles were equal and opposite but unknown

$(1)$.Then meaning the small triangle in the core must be an equilateral triangle, true but no helpful maths here, say $x $ is the length we are looking for

$(2)$.Or maybe the angle of the right vertex as is divided into $3$ parts, $\theta_{1}, \theta_{2}, \theta_{3}$ , the sub angles share eachother $\theta_{2} = 2 \theta_{1}$ and $\theta_{3} = 3 \theta_{1}$, so that the angle there is $6 \theta_{1}$, now If we assume the whole shape was an equilateral here, it contradicts $$ A^2+A^2-2AA \cos(6 \theta ) = 3+x+2$$ $$ 2A^2-2A^2 \cos( 6 \theta ) = 5+x$$ $$ 2A^2( 1- \cos( 6 \theta )) = 5+x$$ $ A \ne 5+x$ , $ A > 5+x$,

If we assume the shape was not an equilateral triangle, and that condition holds, then $$ A^2+B^2-2AB \cos(6 \theta ) = 3+x+2$$ In this case, remember the sector formula, which we can also use to approximate triangle $\theta \rightarrow 0$, $ s ≈ r \cdot \theta$, then $ x = B \cdot \theta $, $B$ is the two length the bisects the angle $6 \theta$ $$ B^2+B^2-2BB \cos( \theta ) = x$$ $$2B^2( 1- \cos( \theta)) = x$$ $$ A^2+B^2-2AB \cos(3 \theta ) = 3$$ $$ B^2+C^2-2BC \cos(2 \theta ) = 2$$ $C$ is the last side, now it's easy to see why $x = 1$

(3).Or maybe the whole shape was an equilateral triangle without the angle of that vertex corresponding, check @Brightstart solution

Answer 3

When you see concurrent lines, Ceva's Theorem and Menelaus's Theorem are often the intended way.

Consider the grey triangle, applying Ceva's Theorem yields $$\frac{3}{x}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$

Again, consider the grey triangle with the black transversal line, applying Menelaus's Theorem yields $$\frac{5+x}{2}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$

Equating both equations gives $x=1$.

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PS: I'm not sure why using other lines and triangles (for instance, the line between the line segments with lengths $3$ and $x$) doesn't work. If anyone would like to expand on this answer, please do. Thanks in advance!