Three identical 30-60-90 right triangles are arranged as shown in the figure.

What is the ratio of the $\color{#D61F06}\text{red segment's}$ length to the $\color{#3D99F6}\text{blue segment's}$ length?

The answer is 3.5.

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Very efficient method, avoids trigonometry ....

Arunava Das
- 3 years, 5 months ago

Minor typo, in equation (1) the radical in the y-intercept should be root 3, not root 2. Got me all confused for a bit where the root 2 came from.

James Sievers
- 3 years, 4 months ago

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Thanks, I've corrected it! Sorry for time you lost recalculating.

Uros Stojkovic
- 3 years, 4 months ago

*
Relevant wiki:
Cosine Rule (Law of Cosines)
*

We expand the graph into a complete hexagon, and found that the six linked segment construct another smaller inside hexagon.

Let $AO = 2$ so then $OB = 1$ , using Law of Cosine obtain $AB = \sqrt{7}$ .

Using different area expressions of $\Delta AOB$ , yield

$AO \times OB \sin{120^\circ} = AB \times OC$ then get $OC = \sqrt{\frac{3}{7}}$ .

Let's observed closer at the smaller hexagon, we have $OD \cos{30^\circ} = OC$ and gives us $OD = \frac{2}{\sqrt{7}}$ .

Finally, the ratio of the red segment's length to the blue segment's length will be $\frac{AB}{OD} = \boxed{\frac{7}{2}}$ .

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Wow, it is really interesting to notice how constructing the Hexagon makes the job much simpler.

Agnishom Chattopadhyay
- 3 years, 5 months ago

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Ya, I knew using coordinate is intuitive, but I want to find a beautiful way to do this.

Kelvin Hong
- 3 years, 5 months ago

Why OB is 1? What is the relation with AO?

Michael Szerman
- 3 years, 4 months ago

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It is a arbitrary value, I just let OA = 2 and OB = 1 base on the relationship of $sin {30^\circ}$ . Because question is asking about the ratio, so the values of OA and OB don't matter.

Kelvin Hong
- 3 years, 4 months ago

(my English is poor) i use the axis to solve it ,may be using the geometric is more good)

my method:
1:calculate AO (use AB and BO)

2:determine the equation of AO

3:determine the equation of DE

4:determine the coordinate of D and E

5: calculate the length of DE

END

|3 is the square root of 3

- assume EO=1 BO=|3

so AC=|3 EC=1

EB=2

so cb=1

∠ACB = 90

so AB=2 and ∠ABO=90

A(2，|3)

AO=|7

AO is y=
$\frac{|3}{2}$
x

C is the midpoint of EB

C (
$\frac{1}{2}$
,
$\frac{|3}{2}$
）

F(3,0)

so FC is y=-
$\frac{|3}{5}$
x +
$\frac{33}{5}$

then D (
$\frac{6}{7}$
,
$\frac{3|3}{7}$
）

DE=
$\frac{2|7}{5}$

the Ans is 3.5

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Interesting problem. I have used similar method, but made silly mistakes.Here my solutions: We can solve the problem by coordinate geometry

Let the origin is at the point E(0,0).Wlog let the triangle has sides of 1,√3,2. So the coordinates of point O(1,0), B(1,√3), C(1/2,1/2 √3),A(-1, √3), and F(-2,0). The equations of line CF: y=√3/5 (x- 2), AO: y = -√3/2 (x- 1), So we can find the intersection point of these two lines (1/7 , 3√3/7) So the ratio of the red line to the green line is 7√7/√28 = 3.5

rab gani
- 3 years, 5 months ago

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Why is triangle ADF equilateral?

Valeri Milanov
- 3 years, 4 months ago

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I expected such a question! On the segment $BC$ we determine the point $D$ , so that $AF=AD$ . The rotation of the triangle on the right side around $A$ is by the angle of $\dfrac\pi3$ , whereby $AD$ maps into $AF$ , accordingly $FD=AF=AD$ .

A Former Brilliant Member
- 3 years, 4 months ago

Sorry，I'm just not good at English.So I decided to draw a picture to show the solution.

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TLDR;

${\color{#D61F06} R}=\sqrt{2.5^{2}+{\frac{\sqrt{3}}{2}}^{2}} = \sqrt{7}$

$\Delta AEC similar \Delta ACD$

$\frac{\sqrt{7}}{2}*\frac{\sqrt{7}}{1}=3.5$

I thought this was a rather elegant solution, but I could not find a way to prove $\Delta AEC$ is similar to $\Delta ACD$ in a simple way

- Since all three 30-60-90 triangles are the same the red line is the same as the black line. $\Delta AEC = \Delta FBC$
- The triangle ABC we know because it is a 30-60-90 triangle its sides are $AB=\sqrt{3}$ , $AC=2$ , and $BC=1$ (or any multiple) segment $BC=1$ and segment $GC=2$ . We know $BC$ is half way between segment $CE$ and segment $GE$ so the dimensions of $\Delta BFH$ are $2.5,\sqrt{3}/2$
- Solve for the black and red line $\Delta AEC = \Delta FBC$ ${\color{#D61F06} R}=\sqrt{2.5^{2}+{\frac{\sqrt{3}}{2}}^{2}} = \sqrt{7}$ Now we know $\Delta AEC$ 's sides $AC=2,CE=1,AE=\sqrt{7}$
- We can determine $\Delta AEC$ is similar to $\Delta ACD$ Recognizing $\angle FDE$ is the interior angle of a regular hexagon $\angle FDE=120$ ToDo (please help)
- Given $\Delta AEC$ is similar to $\Delta ACD$ we set up the proportion of the larger triangle $AE$ to the smaller triangle $AC$ and the proportion of the longest side $AE$ to the shortest side $CE$ .

$\frac{\sqrt{7}}{2}*\frac{\sqrt{7}}{1}=3.5$

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You could do the same with Geogebra too, right?

Agnishom Chattopadhyay
- 3 years, 5 months ago

Wouldn't it be easier to just get a ruler and measure the ratio of the 2 lines on your screen directly?

Pi Han Goh
- 3 years, 5 months ago

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I measured with my fingers and guessed right on the second try.

Rhonda Sobon
- 3 years, 5 months ago

the blue segment can be seen a side of the equilateral triangle that lies on the red segment

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${\color{#D61F06} R_{length}} = \sqrt{2^{2} + \sqrt{3}^{2}} = \sqrt{7}$ Set up the equations: $\begin{aligned} & {\color{#D61F06} (1)}~y = -\frac{\sqrt{3}}{2}x + \frac{3\sqrt{3}}{2} \\ & {\color{#333333} (2)}~y = \frac{\sqrt{3}}{5}x. \end{aligned}$ Find the point of intersection: $\begin{aligned} \left ( \frac{\sqrt{3}}{5} + \frac{\sqrt{3}}{2} \right )x &= \frac{3\sqrt{3}}{2} \\ \frac{7\sqrt{3}}{10}x & = \frac{3\sqrt{3}}{2} \\ \frac{7}{5}x & = 3 \\ x &= \frac{15}{7} \\ \implies y &= \frac{3\sqrt{3}}{7}.\end{aligned}$ Now, you have: ${\color{#3D99F6} B_{length}} = \sqrt{\left ( \frac{15}{7}-2 \right )^{2} + \left ( \frac{3\sqrt{3}}{7}\right )^{2}} =\sqrt{\frac{28}{49}} = \frac{2\sqrt{7}}{7}.$ Hence, $\frac{{\color{#D61F06} R_{length}}}{{\color{#3D99F6} B_{length}}} = \dfrac{\sqrt{7}}{\frac{2\sqrt{7}}{7}} = \frac{7}{2} = \boxed{3.5}.$