Are you smarter than me? 44

Geometry Level 5

Let A B C D \color{#3D99F6}{ABCD} be a quadrilateral in which A B ∥ C D \color{#EC7300}{AB ~\parallel~ CD} and ⊥ \perp to A D \color{#69047E}{AD} and A B = 3 C D \color{#EC7300}{AB=3CD} And the area of quadrilateral is 4 \color{#3D99F6}{4} .If a circle can be drawn touching all the sides of the quadrilateral,find its radius


The answer is 0.8660.

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Mehul Chaturvedi by Mehul Chaturvedi , 22, India

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5 solutions

Aditya Kumar
Dec 26, 2014

Let E E on A B AB be such that C E CE ⊥ \perp A B AB . Take A B = 3 x AB=3x and A D = y AD=y .

  1. Area of quadrilateral = 2 x y 2xy

  2. C D CD = x x . So, using Pitot's Theorem for tangential quadrilaterals, B C BC = 4 x − y 4x-y . Also note that C E = y CE = y and E B = 2 x EB = 2x .

  3. Apply Pythagoras' theorem on triangle BEC and then : y 2 + ( 2 x ) 2 = ( 4 x − y ) 2 y^2 + (2x)^2 = (4x-y)^2

    implying, on solving: 3 x = 2 y 3x=2y or x = 2 y 3 x=\frac{2y}{3} So, area of quadrilateral = 2 x y = 4 y 2 3 = 4 2xy = \frac{4y^2}{3} = 4 ,

    yielding y = 3 y = \sqrt{3}

  4. A final note is the observation : y = 2 r , r y = 2r, r being the inradius.

Thus r = 3 2 r= \frac{\sqrt{3}}{2}

Although, I am sure I h a v e have seen this problem before elsewhere!

12 Helpful 0 Interesting 0 Brilliant 0 Confused

The problem states that AB= 3CD.

In your solution you let AB = x and you say CD = 3x.

Therefore by your reasoning you are concluding that x = 9x.

Can you explain this?

Guiseppi Butel - 6 years, 5 months ago

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Oh no Guiseppi Butel! My fault. In my notebook, I mislabelled the diagram. All pertinent corrections have been made. Do have a look.

Aditya Kumar - 6 years, 5 months ago

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Your statement 4

"A final note is the observation y = 2r, r being the in radius." is not true.

I challenge you to draw this diagram!

Guiseppi Butel - 6 years, 5 months ago

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@Guiseppi Butel – Do challenge this simple fact (diagram or whatever): " A circle touching two distinct parallel lines must have diameter equal to the distance between the two lines" The incircle must touch AB and CD (parallel segments) so, by the above fact y=2r is guaranteed. I post a picture.

Aditya Kumar - 6 years, 5 months ago

Yeas it appeared in RMO 2006

Mehul Chaturvedi - 6 years, 5 months ago

I too did in this way

Nikkil V - 4 years, 5 months ago

I miss the problem because of calculation mistake. But it has a different approach so giving it below.

Please refer to the Fig placed in the coordinate axes.
L e t r b e t h e r e q u i r e d r a d i u s , D C = X , D F t h e ⊥ b i s e c t o r o f A D . ⟹ G ( r , r ) , A D = 2 r , A B = 3 X , G B t h e a n g l e b i s e c t o r o f α . 4 = a r e a = 1 / 2 ∗ ( A B + D C ) ∗ A D = 1 / 2 ∗ 4 X ∗ 2 r = 4 ∗ X ∗ r . ∴ X r = 1 a n d r = 1 / X . . . [ 1 ] . T a n α = s l o p e o f G B = 2 r − 0 3 X − X = r X = r 2 . . . . b y [ 1 ] . . . . . . . [ 2 ] . I d e n t i t y T a n ( α / 2 ) = − 1 + 1 + T a n 2 α T a n α = − 1 + 1 + r 4 r 2 , + t i v e s i g n s i n c e α / 2 > 0........ [ 3 ] B u t s l o p e o f G B = 2 r − r 3 X − r = r 3 X − r = r 2 3 − r 2 . . . . [ 4 ] ∴ [ 3 ] = [ 4 ] . ⟹ − 1 + 1 + r 4 r 2 = r 2 3 − r 2 S q u a r i n g a f t e r a d j u s t m e n t ⟹ ( 1 + r 4 ) ( 3 − r 2 ) 2 = ( r 4 − r 2 + 3 ) 2 ⟹ r 8 − 6 r 6 + 10 r 4 − 6 r 2 + 9 = r 8 − 2 r 6 + 7 r 4 − 6 r 2 + 9 S i n c e r ≠ 0 , r 2 = 3 4 . r = 3 2 . a n g l e S l o p e o f A G = 2 r − r 0 − r = − 1 , ∠ E G A = 135 , s o i t i s a n g l e b i s e c t o r a s r e q u i r e d . S l o p e o f D G = r − 0 r − 0 = 1 , ∠ G D C = 45 , s o i t i s a n g l e b i s e c t o r a s r e q u i r e d . S l o p e o f B G = 2 r − r 3 / r − − r = r 2 3 − − r 2 = 3 / 4 3 − 3 / 4 = 3 1 . S l o p e o f C G = r − 0 r − 1 / r = r 2 r 2 − 1 = 3 / 4 3 / 4 − 1 = − 1 3 . ∴ C G ⊥ B G . ⟹ C G b i s e c t s ∠ D C G ∵ B G b i s e c t s ∠ G B A . Let~ r~ be~ the~ required~ radius,~~~ DC=X,~~~ DF~the~\bot~bisector~of~AD.\\ \implies~~ G(r,r),~ AD=2r,~AB=3X, GB~the~ angle~bisector~of~\alpha.\\ 4=area=1/2*(AB+DC)*AD=1/2*4X*2r=4*X*r.~~~\therefore~Xr=1 ~~~and~~~r=1/X...[1].\\ Tan\alpha=slope~of~GB=\dfrac{2r-0}{3X - X}=\dfrac r X= r^2....by~[1].......[2].\\ Identity~Tan(\alpha/2)=\dfrac{-1+\sqrt{1+Tan^2\alpha} } {Tan\alpha} \\ =\dfrac{-1+\sqrt{1+r^4} }{r^2},~~~+tive~sign~since~\alpha/2>0........[3]\\ But~slope~of~GB=\dfrac{2r-r}{3X-r}=\dfrac r {3X-r}=\dfrac{r^2}{3-r^2}....[4]\\ \therefore~[3] =[4].\\ \implies~~~\dfrac{-1+\sqrt{1+r^4} }{r^2}= \dfrac{r^2}{3-r^2}~~~\\ Squaring~after~adjustment~~\implies~(1+r^4)(3-r^2)^2 =(r^4 -r^2+3)^2 \\ \implies~~r^8-6r^6+10r^4 - 6r^2+9= r^8-2r^6+7r^4 - 6r^2+9 \\ Since~r\neq 0,~~r^2=\dfrac 3 4 .~~~~~\Large \color{#D61F06}{r=\dfrac {\sqrt3}{2}}. angle\\ Slope~of ~AG =\dfrac{2r-r}{0-r}=-1,~\angle~EGA=135,~so~it~is~angle~bisector~as ~required.\\ Slope~of ~DG =\dfrac{r-0}{r-0}=1,~\angle~GDC=45,~so~it~is~angle~bisector~as ~required.\\ Slope~of ~BG =\dfrac{2r-r}{3/r--r}=\dfrac{r^2}{3--r^2}=\dfrac{3/4}{3-3/4}=\dfrac 3 1.\\ Slope~of ~CG =\dfrac{r-0}{r-1/r}=\dfrac{r^2}{r^2-1}=\dfrac{3/4}{3/4-1}=-\dfrac 1 3.\\ \therefore~CG~\bot~BG.~~\implies~CG~bisects~\angle~DCG~\because~BG~bisects~\angle~GBA.

Niranjan Khanderia - 4 years, 2 months ago

i didnot get how did u take ad perpendicular to dc... plz explain

Amit Patel - 3 years, 10 months ago
Chew-Seong Cheong
Dec 27, 2014

Let C D = a CD = a and the radius of the circle be r r , then the area of the quadrilateral is:

A = 2 r ˙ a + 3 a 2 = 4 a r = 4 ⇒ a = 1 r \quad A = 2r\dot{} \dfrac {a+3a}{2} = 4ar = 4\quad \Rightarrow a = \dfrac {1}{r}

Now let ∠ A B C = θ \angle ABC = \theta , then tan ⁡ θ = 2 r 2 a = r a = r 2 \tan {\theta} = \dfrac {2r}{2a} = \dfrac {r}{a} = r^2

We note that: tan ⁡ θ 2 = r 3 a − r = r 2 3 − r 2 \tan {\frac {\theta}{2}} = \dfrac {r}{3a-r} = \dfrac {r^2}{3-r^2} .

Since tan ⁡ θ = 2 tan ⁡ θ 2 1 − tan ⁡ 2 θ 2 ⇒ r 2 = 2 r 2 3 − r 2 1 − ( r 2 3 − r 2 ) 2 \tan {\theta} = \dfrac {2\tan {\frac {\theta}{2}}}{1 - \tan^2 {\frac {\theta}{2}}} \quad \Rightarrow r^2 = \dfrac {\dfrac {2r^2}{3-r^2}} {1-\left( \dfrac {r^2}{3-r^2 } \right)^2 }

⇒ 1 − ( r 2 3 − r 2 ) 2 = 2 3 − r 2 ⇒ ( 3 − r 2 ) 2 − r 4 = 2 ( 3 − r 2 ) \quad \Rightarrow 1-\left( \dfrac {r^2}{3-r^2 } \right)^2 = \dfrac {2}{3-r^2} \quad \Rightarrow (3-r^2)^2 - r^4 = 2(3-r^2)

⇒ 9 − 6 r 2 + r 4 − r 4 = 6 − 2 r 2 ⇒ 4 r 2 = 3 ⇒ r = 3 2 \quad \Rightarrow 9-6r^2+r^4-r^4 = 6-2r^2 \quad \Rightarrow 4r^2 = 3 \quad \Rightarrow r = \boxed {\frac {\sqrt{3}}{2}}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Shouldn't it be tan ⁡ θ = 2 r 3 a \tan{\theta} = \frac{2r}{3a} and not 2 r 2 a \frac{2r}{2a} as you claim ?

Magne Myhren - 6 years, 5 months ago

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It should be tan ⁡ θ = 2 r 2 a \tan {\theta} = \frac {2r}{2a} . See the diagram postd by Aditya Kumar. He used x x instead of a a .

Chew-Seong Cheong - 6 years, 5 months ago
Kartik Sharma
Dec 28, 2014

Well, this is how I solved it without any trigonometry.

Let CD be x and AD be y.

We will construct CE parallel AD. Now, CEAD is a rectangle and hence, area of rectangle CEAD = x y xy

Area of BEC = 1 2 × 2 x × y \frac{1}{2} \times 2x \times y

a r ( B E C ) + a r ( C E A D ) = a r ( A B C D ) ar(BEC) + ar(CEAD) = ar(ABCD)

1 2 × 2 x × y + x y = 4 \frac{1}{2} \times 2x \times y + xy = 4

x y = 2 xy = 2 , y = 2 x y = \frac{2}{x}

In BEC, using Pythagoras Theorem,

2 x 2 + 4 x 2 = B C \sqrt{{\frac{2}{x}}^{2} + 4{x}^{2}} = BC

As there can be an incircle,

AB + CD = AD + BC

4 x = 2 x + 2 x 1 + x 4 4x = \frac{2}{x} + \frac{2}{x}\sqrt{1 + {x}^{4}}

2 x 2 − 1 = 1 + x 4 2{x}^{2} - 1 = \sqrt{1 + {x}^{4}}

4 x 4 + 1 − 4 x 2 = 1 + x 4 4{x}^{4} + 1 - 4{x}^{2} = 1 + {x}^{4}

x 2 = 4 3 {x}^{2} = \frac{4}{3}

x = 2 3 x = \frac{2}{\sqrt{3}}

Also, A r e a = r s Area = rs

4 = r ( 8 x 2 ) 4 = r(\frac{8x}{2})

r = 3 2 ≈ 0.866 r = \frac{\sqrt{3}}{2} \approx 0.866

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Nihar Mahajan
Dec 28, 2014

this question is from crmo 2006! answer is 3^(1/2)/2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for uploading your hand written solution!

Calvin Lin Staff - 6 years, 5 months ago
Mohanish Mayank
Dec 26, 2014

let CD=x ; and radius be r join centre of circle O to C and B which will be the angle bisectors; draw perpendicular from O to CD and AB since /CBA+/BCD=180 it implies sum of /OCD and /OBA IS 90 ; from eqating area we will get rx=1 .............i

now tanOCD=r/x-r and tanOBA=r/3x-r since /OCD+/OBA=90; it implies tanOCD×tanOBA=1......ii

solving both eqation r can be found

0 Helpful 0 Interesting 0 Brilliant 0 Confused

With that radius the circle touches only 3 of the 4 sides.

Guiseppi Butel - 6 years, 5 months ago

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since i m saying that joining centre of circle to B and C will be angle bisector it implies that it will touch fourth side also

there is not any incorrectness in this question think calmly

Mohanish Mayank - 6 years, 5 months ago

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Circle in Quad Circle in Quad

Guiseppi Butel - 6 years, 5 months ago

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@Guiseppi Butel – Such a trapezium exists. You can do the calculations that if it has a height of 3 \sqrt{3} , and base lengths of 2 3 3 \frac{2\sqrt{3} } { 3} and 2 3 2 \sqrt{3} , then the last side will have length of 5 3 3 \frac {5 \sqrt{3} } {3} . You can set up the coordinate geom, to show that there is a circle with radius 3 2 \frac{ \sqrt{3} } {2} that is tangential to all 4 sides.

We have the following theorem (proof not provided):

The sum of opposite sides of a convex quadrilaterial are equal if and only if we can have a circle that is inscribed within.

Calvin Lin Staff - 6 years, 5 months ago

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@Calvin Lin – I finally agree. My disputes were with two different statements provided in the solutions given; one that gave a conclusion that resulted in x = 9x, and a statement that " that joining centre of circle to B and C will be angle bisector it implies that it will touch fourth side also"

Here is my solution: My Soln My Soln

Guiseppi Butel - 6 years, 5 months ago

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@Guiseppi Butel – Right. I agree that there is no substantiation in the solution which explain why such a geometric configuration must occur. Assuming that a solution exists, that must be the solution. For completeness, it is good practice to check that such a configuration actually exists, i.e. that the solution is indeed valid.

Calvin Lin Staff - 6 years, 5 months ago

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@Calvin Lin – in my solution i use rx=1 which shows circle is touching the parallel sides i use tanOCD=r/r-x which shows it touches perpendicular side and the angle bisector thing which shows it touches slant side also

there is no incompleteness in the solution!

Mohanish Mayank - 6 years, 5 months ago

@Guiseppi Butel – You have not got the correct trapezium...how can you expect the poor incircle to fit in?! I can easily see the flaw. Had you taken: A D = 3 AD = \sqrt{3} , A B = 2 3 AB = 2\sqrt{3} , C D = 2 3 CD = \frac{2}{\sqrt{3}} then you would have gotten it.

I can see clearly your A B AB is less than double of A D AD , when a double is expected.

Do go through my construction algorithm posted.

Aditya Kumar - 6 years, 5 months ago

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