Individual Integral Triangle

Geometry Level 4

Given a circumradius 9 9 and the inradius 4 4 , there exist some triangles with an integer area A A . Find A A .


The answer is 90.

Michael Huang by Michael Huang , 29, USA

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1 solution

Chris Lewis
Apr 14, 2021

Notation: R R is the circumradius; r r the inradius); O O the circumcentre; I I the incentre; A , B , C A,B,C are the vertices of the triangle (and also the angles at those vertices); a , b , c a,b,c are the lengths of the sides opposite those vertices; T T is the area of the triangle.

By Euler's theorem on triangles, the circumcentre and the incentre are a distance d = R 2 − 2 R r d=\sqrt{R^2-2Rr} apart.

Take O O as an origin; let A A have coordinates ( R , 0 ) (R,0) and I ( d cos ⁡ ( θ ) , d sin ⁡ ( θ ) ) I(d \cos(\theta),d \sin(\theta)) . Then all the different triangles satisfying the conditions are found by varying θ \theta .

Since the incentre lies on the angle bisectors, ∠ I A B = A 2 \angle IAB=\frac{A}{2} ; so sin ⁡ A 2 = r A I \sin \frac{A}{2} = \frac{r}{AI}

By the extended sine rule, a = 2 R sin ⁡ A a=2R \sin A .

Now consider the triangle B I C BIC . By definition of the incentre, the altitude from I I has length r r . Also it's easy to show that ∠ B I C = A + π 2 \angle BIC = \frac{A+\pi}{2} . Again using the fact the incentre lies on the angle bisectors, ∠ I B C = B 2 \angle IBC=\frac{B}{2} .

Combining all of this, B C = a = 2 R sin ⁡ A = r ( cot ⁡ B 2 + cot ⁡ C 2 ) = r ( cot ⁡ B 2 + tan ⁡ A + B 2 ) BC=a=2R\sin A=r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)=r\left(\cot \frac{B}{2}+\tan \frac{A+B}{2}\right)

This can be written as a quadratic in cot ⁡ B 2 \cot\frac{B}{2} .

After a LOT of trig (please comment if there's a shortcut to this), we get b c = 4 r R + A I 2 bc=4rR + AI^2

The area is given by T = 1 2 b c sin ⁡ A = r ( 4 r R A I 2 + 1 ) A I 2 − r 2 T=\frac12 bc \sin A = r\left(\frac{4rR}{AI^2} + 1\right) \sqrt{AI^2-r^2}

Now, we can write all this in terms of θ \theta , or note that A I 2 AI^2 varies from R − d R-d to R + d R+d . Letting x = A I 2 x=AI^2 , we have T = r ( 4 r R x + 1 ) x − r 2 T=r\left(\frac{4rR}{x} + 1\right) \sqrt{x-r^2}

Differentiating, d T d x = r ( 8 r 3 R − 4 r R x + x 2 ) 2 x 2 x − r 2 \frac{dT}{dx}=\frac{r\left(8r^3 R-4rRx+x^2 \right)}{2x^2 \sqrt{x-r^2}}

Setting the derivative equal to zero, 8 r 3 R − 4 r R x + x 2 = 0 8r^3 R-4rRx+x^2=0

With R = 9 R=9 and r = 4 r=4 , this is x 2 − 144 x + 4608 = 0 x^2-144x+4608=0

with roots x = 48 x=48 and x = 96 x=96 . The corresponding areas are T max = 64 2 ≈ 90.51 T_\text{max}=64\sqrt2\approx90.51 and T min = 40 5 ≈ 89.44 T_\text{min}=40\sqrt5\approx89.44

Since there is just one integer in the interval [ T min , T max ] [T_\text{min},T_\text{max}] , this is the answer we're looking for, namely 90 \boxed{90} .

In fact, we can solve this in general; we get T min = r ( 3 R + d ) r ( 2 R − r + 2 d ) R + d T_\text{min} = \frac{r (3 R + d) \sqrt{r (2R -r+ 2 d)}}{R + d} and T max = r ( 3 R − d ) r ( 2 R − r − 2 d ) R − d T_\text{max} = \frac{r (3 R - d) \sqrt{r (2R -r- 2 d)}}{R - d}

1 Helpful 1 Interesting 4 Brilliant 0 Confused

Can you find a triangle with circumradius, inradius and area of 9, 4 and 90, respectively?

Pi Han Goh - 1 month, 4 weeks ago

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Good point, it's not really complete without that (although I suppose continuity is obvious so we can use the intermediate value theorem).

With R = 9 R=9 , r = 4 r=4 , T = 90 T=90 , in the area formula above we get 90 = 4 ( 144 x + 1 ) x − 16 90=4\left(\frac{144}{x}+1\right)\sqrt{x-16}

Squaring, multiplying through, tidying, we get (the slightly ridiculous) 4 x 3 − 937 x 2 + 64512 x − 1327104 = 0 4 x^3-937x^2 + 64512 x -1327104 = 0

This has three real roots (as we might expect - the triangle could be rotated/reflected, changing its position without changing its size). They're not nice, though! Feel free to find the exact forms.

The smallest root is around x = 38.901 x=38.901

Recall x = A I 2 x=AI^2 ; so we have A I AI , from which we can work out the side a a ; we get a ≈ 17.7145 a\approx 17.7145

Rather than go through the whole process of finding b , c b,c via trig, we can actually just find out what sidelengths we get with the different roots of the cubic. These turn out to be around 15.4382 15.4382 and 11.8473 11.8473 ; and you can check that the triangle with these three sides works.

I'm a bit surprised to see that (up to reflection) this triangle is unique. @Pi Han Goh - you're normally very good at dealing with cubics, is there a way of writing these sidelengths exactly?

Chris Lewis - 1 month, 4 weeks ago

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Let z = 1 3 tan ⁡ − 1 ( 8 9 23 3 ) , z = \dfrac13 \tan^{-1} \left( \dfrac89 \sqrt{ \dfrac{23}3} \right) , then the side lengths of this triangle are:

  • 15 − 35 3 cos ⁡ ( z ) ≈ 11.8473 15 - \sqrt{\dfrac{35}3} \cos (z) \approx 11.8473 ,

  • 15 + 105 6 cos ⁡ ( z ) − 35 2 sin ⁡ ( z ) ≈ 15.438 , 15 + \dfrac{\sqrt{105}}{6} \cos(z) - \dfrac{\sqrt{35}}{2} \sin(z) \approx 15.438, and

  • 15 + 35 12 cos ⁡ ( z ) + 35 2 sin ⁡ ( z ) ≈ 17.714 15 + \sqrt{\dfrac{35}{12}} \cos(z) + \dfrac{\sqrt{35}}{2} \sin(z) \approx 17.714

How did I do it?

Because sin ⁡ ( A 2 ) = r A I = 4 x ⇒ A = 2 sin ⁡ − 1 ( 4 x ) \sin \left(\dfrac A2 \right) = \frac r{AI} = \dfrac 4{\sqrt x} \quad \Rightarrow \quad A = 2 \sin^{-1} \left( \dfrac 4 {\sqrt{x}} \right)

Then, a = 2 R sin ⁡ ( A ) = 2 ⋅ 9 ⋅ sin ⁡ [ 2 sin ⁡ − 1 ( 4 x ) ] = 144 x − 16 x a = 2R \sin(A) = 2\cdot9 \cdot \sin \left [2 \sin^{-1} \left( \dfrac 4 {\sqrt{x}} \right) \right ] = \dfrac{144\sqrt{x-16}}{x}

Thus, if 4 x 3 − 937 x 2 + 64512 x − 1327104 = 0 , 4x^3 -937x^2 +64512x-1327104 = 0 , then the side lengths of this triangle are the three possible values of 144 x − 16 x . \dfrac{144\sqrt{x-16}}{x}.

Pi Han Goh - 1 month, 4 weeks ago

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@Pi Han Goh – Ugh, where's the "brilliant" button for comments when you need it? That's very clever indeed.

I'm really interested also in whether there is some obvious reason why b c = 4 r R + A I 2 bc=4rR+AI^2

I spent way too long working that out - mainly because Wolfram|Alpha told me to take my page-long equations elsewhere - but it's such a neat expression it'd be nice to explain it more intuitively. Any thoughts?

By the way, I've just added in some formulas for the min/max areas in general. It turns out these work out fairly nicely too.

Chris Lewis - 1 month, 4 weeks ago

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@Chris Lewis – From here , R = a b c 4 r s ⇔ 4 r R = a b c s = 2 a b c a + b + c . (1) R = \dfrac{abc}{4rs} \quad \Leftrightarrow \quad 4rR = \dfrac{abc}{s} = \dfrac{2abc}{a + b + c} .\tag1

And from here , A I = c ⋅ sin ⁡ ( B / 2 ) cos ⁡ ( C / 2 ) ⇔ ( A I ) 2 = c 2 ⋅ 2 sin ⁡ 2 ( B / 2 ) 2 cos ⁡ 2 ( C / 2 ) = c 2 ⋅ 1 − cos ⁡ ( B ) 1 + cos ⁡ ( C ) (2) AI = c\cdot \dfrac{ \sin(B/2)}{\cos(C/2)} \quad \Leftrightarrow \quad (AI)^2 = c^2 \cdot \dfrac{ 2\sin^2(B/2)}{2\cos^2(C/2)} = c^2 \cdot \dfrac{1- \cos (B)}{1 + \cos(C)}\tag2

Using cosine rule , cos ⁡ ( B ) = a 2 + c 2 − b 2 2 a c , cos ⁡ ( C ) = a 2 + b 2 − c 2 2 a b (3) \cos(B) = \dfrac{a^2 + c^2-b^2}{2ac}, \quad\quad\quad \cos(C) = \dfrac{a^2 + b^2 - c^2}{2ab }\tag3

Substituting ( 1 ) , ( 2 ) (1), (2) and ( 3 ) (3) into 4 r R + ( A I ) 2 4rR + (AI)^2 simplifies to b c . bc.

Pi Han Goh - 1 month, 4 weeks ago

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@Pi Han Goh – Wow, fantastic, good searching. Definitely looks like I took the long way round. It's interesting how powerful half-angle formulas are here (I'm used to using them in calculus problems, but not so much with triangles).

Chris Lewis - 1 month, 4 weeks ago

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