Find x x

Geometry Level 3


The answer is 12.7017.

Aly Ahmed by Aly Ahmed , 34, Egypt

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4 solutions

7 Helpful 5 Interesting 9 Brilliant 0 Confused

can u help.. https://brilliant.org/discussions/thread/how-to-approach-this-question/

Syed Hamza Khalid - 1 year, 4 months ago

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you are asking me?

nibedan mukherjee - 1 year, 4 months ago

Let the radius of the circle be r r . Then cos ⁡ 30 ° = √ 3 2 = 100 + x 2 − r 2 20 x = 144 + x 2 − r 2 24 x \cos {30\degree}=\dfrac{√3}{2}=\dfrac{100+x^2-r^2}{20x}=\dfrac{144+x^2-r^2}{24x} . Or 2 √ 3 x = 44 2√3x=44 or x = 22 √ 3 = 12.7017... x=\dfrac{22}{√3}=\boxed {12.7017...}

1 Helpful 2 Interesting 7 Brilliant 3 Confused

Refer to the diagram of Chew-Seong Cheong. Let O be the centre of the circle. Note that / DOC=60°. Also / BOC=60° which make Tr. DOC & Tr. BOC equilateral and hence DC=BC=r. Then it follows from Tr. ADC that: √3/2 = ​(100+x²-r²)/(20x). Likewise, from Tr. ABC: √3/2 = ​(144+x²-r²)/(24x). Solve the two equations to get x=22/√3. Nice solution Alak Bhattacharya.

ajit athle - 1 year, 4 months ago

Providing a general solution as suggested by @Nikola Alfredi . Therefore, instead of 10 10 , 12 12 , and 3 0 ∘ 30^\circ , we have a a , b b , and θ \theta respectively.

Since A B C D ABCD is a cyclic quadrilateral , then ∠ B A C = ∠ B D C = θ \angle BAC = \angle BDC = \theta and ∠ C A D = ∠ C B D = θ \angle CAD = \angle CBD = \theta . Then △ B C D \triangle BCD is isosceles and B C = C D = c BC=CD = c . By Ptolemy's theorem , we have:

A C ⋅ B D = A D ⋅ B C + A B ⋅ D C Let B D = y x y = a c + b c Note that y = 2 c cos ⁡ θ ⟹ x = a c + b c 2 c cos ⁡ θ = a + b 2 cos ⁡ θ \begin{aligned} AC \cdot \blue{BD} & = AD \cdot BC + AB \cdot DC & \small \blue{\text{Let }BD=y} \\ x\blue y & = ac+bc & \small \blue{\text{Note that }y = 2c\cos \theta} \\ \implies x & = \frac {ac+bc}{2c\cos \theta} = \frac {a+b}{2\cos \theta} \end{aligned}

For a = 10 a=10 , b = 12 b=12 , and θ = 3 0 ∘ \theta = 30^\circ , x = 22 3 ≈ 12.702 x = \dfrac {22}{\sqrt 3} \approx \boxed{12.702} .

2 Helpful 2 Interesting 3 Brilliant 0 Confused

General Formula for such type of questions where

a ≠ b T h e n x = a + b 2 cos ⁡ θ \ \ \displaystyle a \neq b \ \ \ Then \ \ \ x = \frac {a + b}{2 \cos \theta}

Nikola Alfredi - 1 year, 4 months ago

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Thanks for your suggestion. I have provided the general solution.

Chew-Seong Cheong - 1 year, 4 months ago

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You're welcome

Nikola Alfredi - 1 year, 4 months ago

@Nikola Alfredi ,Suppose the angle between a and b are different, Then what would be your answer? Is it a + b cos ⁡ α + cos ⁡ β ? \frac{a+b}{\cos{\alpha}+\cos{\beta}}?

Winod DHAMNEKAR - 1 year, 4 months ago

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@Winod DHAMNEKAR

I will tell you soon but I belief that the question will be incomplete with the given data. It must require at least one more value point...

Nikola Alfredi - 1 year, 4 months ago

@Winod DHAMNEKAR

Hi, As I said there would be no question like this but if it exists then solution would be :

Assume ∠ B A C = β \angle BAC = \beta and ∠ D A C = α \ \angle DAC = \alpha

Solution : (A polynomial)

2 x 4 cos ⁡ ( α + β ) − 4 x 3 ( a cos ⁡ α + b cos ⁡ β ) ( cos ⁡ ( α + β ) ) + 2 x 2 [ ( a 2 + b 2 + 4 a b cos ⁡ α cos ⁡ β ) ( cos ⁡ ( α + β ) ) + 1 ] + 2 x [ 2 a b ( b cos ⁡ α + a cos ⁡ β ) cos ⁡ ( α + β ) − ( a cos ⁡ α + b cos ⁡ β ) ] + 2 a b cos ⁡ ( α + β ) + a 2 b 2 = 0 \displaystyle 2x^4 \cos(\alpha + \beta) - 4x^3 (a \cos \alpha + b \cos \beta)(\cos(\alpha + \beta)) + 2x^2[(a^2 +b^2 + 4ab \cos \alpha \cos \beta)(\cos(\alpha + \beta)) + 1] +2x[2ab(b \cos \alpha + a \cos \beta) \cos(\alpha + \beta) - (a \cos \alpha + b \cos \beta)] + 2ab \cos(\alpha + \beta) + a^2b^2 = 0

Nikola Alfredi - 1 year, 4 months ago

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This equation don't give answer for the case a = 10 , b = 12 , α = β = 30 ° a = 10\,,\, b = 12\,,\, \alpha = \beta = 30\degree For this case this equation has no solution.

Shikhar Srivastava - 1 year, 4 months ago

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x 2 = a 2 + b 2 − 2 a b cos ( α + β ) sin 2 ( α + β ) cos 2 ( θ − β ) \large x^2 = \Large \frac{a^2 + b^2 -2ab\text{cos}(\alpha + \beta)}{\text{sin}^2(\alpha + \beta)} \text{cos}^2(\theta - \beta) where

cos 2 ( θ ) = b 2 sin 2 ( α + β ) a 2 + b 2 − 2 a b cos ( α + β ) \Large \text{cos}^2(\theta) = \frac{b^2\text{sin}^2(\alpha + \beta)}{a^2 + b^2 -2ab \text{cos}(\alpha + \beta)}

Shikhar Srivastava - 1 year, 4 months ago

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@Shikhar Srivastava – Ok, thank you

Nikola Alfredi - 1 year, 4 months ago

@Shikhar Srivastava – @Shikhar Srivastava I found the formula just by using Cosine-Rule in Δ B A C & Δ D A C & Δ B C D \Delta BAC \ \& \ \Delta DAC \ \& \ \Delta BCD why does it goes wrong any clue....?

Nikola Alfredi - 1 year, 4 months ago

@Shikhar Srivastava – How did you find it ... just give hint.

Nikola Alfredi - 1 year, 4 months ago

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@Nikola Alfredi – Let O O be the center of the circle. First find the radius of circle. Suppose ∠ A B O = θ . \angle ABO = \theta . Then find expression for θ \theta . Then find ∠ A O C \angle AOC in terms of θ \theta and β \beta . After this in △ A O C , A O , O C , ∠ A O C \triangle AOC\,,\, AO , OC \,,\, \angle AOC is known. you can find A C AC

Shikhar Srivastava - 1 year, 4 months ago

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@Shikhar Srivastava – @Shikhar Srivastava Thank you very much for the help. But did you found why error is coming in my equation......

Nikola Alfredi - 1 year, 4 months ago

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@Nikola Alfredi – What you are doing ? By applying cosine rule what sides or angle are you finding and in which sequence ? Please tell me so that I may help you.

Shikhar Srivastava - 1 year, 4 months ago

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@Shikhar Srivastava – Ok, First I used Cosine rule in Δ A B C \Delta ABC and found B C 2 BC^2 then I Found C D 2 CD^2 , Similarly I found B D 2 BD^2 by Cosine rule and then finally in Δ B C D \Delta BCD I used the Cosine rule.... This is all I did..

I tried your way and I believe it's cool and easy....

Nikola Alfredi - 1 year, 4 months ago

@Shikhar Srivastava – Hey, I got it :)

Nikola Alfredi - 1 year, 4 months ago

@Shikhar Srivastava I think your formula for x 2 x^2 requires some corrections. Put a = 10 , b = 5 , α = 38. 9 ∘ a n d β = 77.2 1 ∘ a=10, b=5,\alpha=38.9^\circ and \beta=77.21^\circ .

Example Link

Winod DHAMNEKAR - 1 year, 4 months ago

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In your link angles α \alpha and β \beta are not taking the entries. Only sides and diagonals are allowed to fill. On putting the values as you said the value is coming x = 14.417 x = 14.417 . Is it not correct ? Please tell me the exact problem so that I will look into it.

Shikhar Srivastava - 1 year, 4 months ago

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In the link, put a=9, b=10,c=5, d=14, diagonal AC=13. You will get β 1 = 38. 9 ∘ , β 2 = 77.2 1 ∘ \beta_1=38.9^\circ, \beta_2=77.21^\circ . You will get diagonal BD(which is x in our case, we want to find out)=14.23. Comparing it to your answer 14.417 ,it is approximately same not exact.

Winod DHAMNEKAR - 1 year, 4 months ago

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@Winod Dhamnekar – The quadrilateral A B C D ABCD in your case is not cyclic.

Shikhar Srivastava - 1 year, 4 months ago

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@Shikhar Srivastava – @Shikhar Srivastava , It is a cyclic quadrilateral. d 1 d 2 = a c + b d , 13 ∗ 14.23 = 9 ∗ 5 + 10 ∗ 14 , ⇒ 184.99 = 185 d_1d_2=ac+bd, 13*14.23=9*5+10*14, \Rightarrow 184.99=185 with only 0.01 absolute error.

Winod DHAMNEKAR - 1 year, 4 months ago

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@Winod Dhamnekar – cos ( B ) = A B 2 + B C 2 − A C 2 2 ⋅ A B ⋅ B C = 100 + 25 − 169 2 ⋅ 10 ⋅ 5 = − 44 100 = − 11 25 \text{cos}(B) = \large\frac{AB^2 + BC^2 - AC^2}{2\cdot AB\cdot BC} = \frac{100 + 25 - 169}{2\cdot 10\cdot 5} = \frac{-44}{100} = \frac{-11}{25}

cos ( D ) = A D 2 + D C 2 − A C 2 2 ⋅ A D ⋅ D C = 81 + 196 − 169 2 ⋅ 9 ⋅ 14 = 108 252 = 3 7 . \text{cos}(D) = \large\frac{AD^2 + DC^2 - AC^2}{2\cdot AD\cdot DC} = \frac{81 + 196 - 169}{2\cdot 9\cdot 14} = \frac{108}{252} = \frac{3}{7}.

For cyclicity of quadrilateral, ∠ B + ∠ D = 180 ° ⇒ ∠ B = 180 ° − ∠ D ⇒ = cos ( B ) = cos ( 180 ° − D ) = − cos ( D ) \angle B + \angle D = 180\degree \Rightarrow \angle B = 180\degree - \angle D \Rightarrow = \text{cos}(B) = \text{cos}(180\degree - D) = -\text{cos}(D) which is not in your case. Hence not cyclic.

Shikhar Srivastava - 1 year, 4 months ago

I got the mistake. The mistake is I forgot to square the term cos ( θ − β ) \text{cos}(\theta - \beta) in the expression of x 2 x^2 . Actually it should be cos 2 ( θ − β ) \text{cos}^2(\theta - \beta) I have corrected it.

Shikhar Srivastava - 1 year, 4 months ago
Vinod Kumar
May 2, 2020

Using property of cyclic quadrilateral, opposite angle sum is 180 degree. Solve 12 sin(x)=10 sin(2 pi/3-x), y=sin(x), z=sin(150-x). Then w=10 z/y =12.7017 is the answer.

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