Trigonometry or Square completing?

Geometry Level 3

In the diagram above, the blue right triangle is partially overlapping the 1 × 1 1\times1 red square, sharing a corner. If A B ‾ = 1 \overline{AB}=1 and B C ‾ = x , \overline{BC}=x, find x x to 3 decimal places.


The answer is 0.883.

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Vitor Santos by Vitor Santos , 21, Brazil

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3 solutions

Andy Hayes
Feb 1, 2017

Define the remaining points in the figure as below:

It can be seen that △ A B C ∼ △ A E F . \triangle ABC \sim \triangle AEF. Then y = 1 x . y=\frac{1}{x}. Apply the Pythagorean Theorem on △ E B D : \triangle EBD:

( x + 1 ) 2 + 1 2 = ( 1 + 1 x ) 2 x 2 + 2 x + 2 = 1 + 2 x + 1 x 2 x 4 + 2 x 3 + x 2 − 2 x − 1 = 0 \begin{aligned} (x+1)^2+1^2 &= \left(1+\frac{1}{x}\right)^2 \\ \\ x^2+2x+2 &= 1+\frac{2}{x}+\frac{1}{x^2} \\ \\ x^4+2x^3+x^2-2x-1 &= 0 \end{aligned}

Edit : See @Mark Hennings comment for how to solve this quartic. The only positive root to the equation above is:

x = 1 2 ( − 1 + 2 + 2 2 − 1 ) ≈ 0.883 . \begin{aligned} x &= \frac{1}{2}\left(-1+\sqrt{2}+\sqrt{2\sqrt{2}-1}\right) \\ \\ &\approx \boxed{0.883}. \end{aligned}

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We can write x 4 + 2 x 3 + x 2 − 2 x − 1 = 0 ( x 2 + x + 1 ) 2 = 2 ( x + 1 ) 2 x 2 + x + 1 = ± 2 ( x + 1 ) x 2 + ( 1 ∓ 2 ) x + ( 1 ∓ 2 ) = 0 ( x + 1 2 ( 1 ∓ 2 ) ) 2 = 1 4 ( 1 ∓ 2 ) 2 − ( 1 ∓ 2 ) = 1 4 ( 1 ∓ 2 ) ( − 3 ∓ 2 ) \begin{aligned} x^4 + 2x^3 + x^2 - 2x - 1 & = 0 \\ (x^2 + x + 1)^2 & = 2(x + 1)^2 \\ x^2 + x + 1 & = \pm\sqrt{2}(x+1) \\ x^2 + (1 \mp \sqrt{2})x + (1 \mp \sqrt{2}) & = 0 \\ \big(x + \tfrac12(1 \mp \sqrt{2})\big)^2 & = \tfrac14(1 \mp \sqrt{2})^2 - (1 \mp \sqrt{2}) \; = \; \tfrac14(1 \mp \sqrt{2})(-3 \mp \sqrt{2}) \end{aligned} For a real root x x , we must choose the sign to make the RHS positive. Thus ( x + 1 2 ( 1 − 2 ) ) 2 = 1 4 ( 2 − 1 ) ( 3 + 2 ) = 1 4 ( 2 2 − 1 ) x + 1 2 ( 1 − 2 ) = ± 1 2 2 2 − 1 \begin{aligned} \big(x + \tfrac12(1 - \sqrt{2})\big)^2 & = \tfrac14(\sqrt{2}-1)(3+\sqrt{2}) \; = \;\tfrac14(2\sqrt{2} - 1) \\ x + \tfrac12(1 - \sqrt{2}) & = \pm\tfrac12\sqrt{2\sqrt{2}-1} \end{aligned} and we now pick the positive root.

A monic quartic can always be written in the form ( x 2 + a x + b ) 2 + c ( x + d ) 2 (x^2 + ax + b)^2 + c(x + d)^2 for some constants a , b , c , d a,b,c,d . Thus a quartic equation can be solved by first taking square roots, and then solving a pair of quadratics. In general you have to be able to solve a cubic equation to determine these four constants, but in this case it was easy.

Mark Hennings - 4 years, 4 months ago

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Awesome, thanks :)

Andy Hayes - 4 years, 4 months ago

You can plot any equation using this tool. https://www.desmos.com/calculator

Vijay Simha - 4 years, 4 months ago

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You also can plot any equation using this tool "Wolfram" (search in Google Play)

I Gede Arya Raditya Parameswara - 4 years, 4 months ago

I tried by taking x= cosδ and converted it into a quadratic in terms of trigonometric terms but I am getting 0.468.I don't know what is wrong with my method.

Dhanvanth Balakrishnan - 4 years, 4 months ago
Raj Mantri
Feb 24, 2017

It can be seen that Triangle BED ~ Triangle BAC

S o , B C / B D = A C / E D So, BC / BD = AC / ED

x/(x+1) = [ √ ( A B ) 2 − ( B C ) 2 ] / 1 [√{(AB)^2 - (BC)^2} ] /1

x = ( x + 1 ) ( √ 1 − x 2 ) x = (x+1)(√1-x^2)

Squaring both sides

x 2 = ( x + 1 ) 2 ( 1 − x 2 ) x^2 = (x+1)^2 (1-x^2)

x 2 = ( x 2 + 2 x + 1 ) ( 1 − x 2 ) x^2 = (x^2+2x+1)(1-x^2)

x 2 = x 2 + 2 x + 1 − x 4 − 2 x 3 − x 2 x^2 = x^2+2x+1- x^4 - 2x^3- x^2

x 2 = 2 x + 1 − x 4 − 2 x 3 x^2 = 2x+1- x^4 - 2x^3

0 = 2 x + 1 − x 4 − 2 x 3 − x 2 0 = 2x + 1 - x^4 - 2x^3 - x^2

Solve this Quartic for x. The only root positive root to above equation is

x = 0.883 x =0.883

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Since △ A B C △ A E F \triangle ABC ~ \triangle AEF it follows that A F = C F = 1 2 AF=CF= \frac{1} {2} which makes the triangles 30-60-90 triangles and so x = 1 2 3 x = \frac{1} {2} \sqrt {3}

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△ A B C \triangle ABC and △ A E F \triangle AEF are similar, not congruent. Your method is good for getting an estimate, since they are so close to being congruent. Note that the exact value is a little bit off:

Estimate: 1 2 3 ≈ 0.866. Actual: 1 2 ( − 1 + 2 + 2 2 − 1 ) ≈ 0.883. \begin{array}{rrl} \text{Estimate: } & \frac{1}{2}\sqrt{3} & \approx 0.866. \\ \text{Actual: } & \frac{1}{2}\left(-1+\sqrt{2}+\sqrt{2\sqrt{2}-1}\right) & \approx 0.883. \end{array}

Andy Hayes - 4 years, 4 months ago

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It did take my answer as correct. But if I look at the picture again my reasoning should lead to a contradiction... So my answer is o deed faulty

Peter van der Linden - 4 years, 4 months ago

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