Just Another 3-4-5 Triangle

Geometry Level 5

Find the area of the non-degenerate triangle with the distances between its vertices and its incenter, I I , equal to 3, 4, and 5.

If K K is the area, submit ⌊ 1 0 4 K ⌋ \lfloor 10^4K\rfloor .


The answer is 198877.

Fletcher Mattox by Fletcher Mattox , 72, USA

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2 solutions

Chris Lewis
Apr 29, 2021

Let the inradius of Δ A B C \Delta ABC be r r . Recall the incentre I I lies on the three angle bisectors of Δ A B C \Delta ABC be r r . By considering Δ A I C \Delta AIC , we have sin ⁡ C 2 = r 3 \sin\frac{C}{2}=\frac{r}{3}

Now, in Δ A I B \Delta AIB , we have ∠ I A B = A 2 \angle IAB=\frac{A}{2} , ∠ I B A = A 2 \angle IBA=\frac{A}{2} and so ∠ A I B = π − A 2 − B 2 = π − A + B 2 = π − π − C 2 = π 2 + C 2 \angle AIB=\pi-\frac{A}{2}-\frac{B}{2}=\pi-\frac{A+B}{2}=\pi-\frac{\pi-C}{2}=\frac{\pi}{2}+\frac{C}{2}

Also, ∠ A I B = cos ⁡ − 1 r 5 + cos ⁡ − 1 r 4 \angle AIB=\cos^{-1} \frac{r}{5}+\cos^{-1} \frac{r}{4}

Putting these together and taking cosines, cos ⁡ ( cos ⁡ − 1 r 5 + cos ⁡ − 1 r 4 ) = cos ⁡ ( π 2 + C 2 ) r 5 ⋅ r 4 − 1 − r 2 5 2 ⋅ 1 − r 2 4 2 = − sin ⁡ C 2 r 2 20 − ( 1 − r 2 25 ) ( 1 − r 2 16 ) = − r 3 \begin{aligned} \cos \left(\cos^{-1} \frac{r}{5}+\cos^{-1} \frac{r}{4}\right) &= \cos \left(\frac{\pi}{2}+\frac{C}{2}\right) \\ \frac{r}{5} \cdot \frac{r}{4} - \sqrt{1-\frac{r^2}{5^2}} \cdot \sqrt{1-\frac{r^2}{4^2}} &=-\sin \frac{C}{2} \\ \frac{r^2}{20} - \sqrt{\left(1-\frac{r^2}{25}\right)\left(1-\frac{r^2}{16}\right)} &=-\frac{r}{3} \end{aligned}

Rearranging and squaring, r 2 20 + r 3 = ( 1 − r 2 25 ) ( 1 − r 2 16 ) r 4 400 + r 3 30 + r 2 9 = 1 − r 2 25 − r 2 16 + r 4 400 r 3 30 + r 2 9 = 1 − r 2 25 − r 2 16 \begin{aligned} \frac{r^2}{20}+\frac{r}{3} &= \sqrt{\left(1-\frac{r^2}{25}\right)\left(1-\frac{r^2}{16}\right)} \\ \frac{r^4}{400}+\frac{r^3}{30}+\frac{r^2}{9} &=1-\frac{r^2}{25}-\frac{r^2}{16}+\frac{r^4}{400} \\ \frac{r^3}{30}+\frac{r^2}{9} &=1-\frac{r^2}{25}-\frac{r^2}{16} \end{aligned}

which after tidying the fractions becomes 120 r 3 + 769 r 2 − 3600 = 0 120r^3+769r^2-3600=0 This has a unique positive root, around r = 1.9002 r=1.9002 . From here, we can work out the sides using c = A B = 5 2 − r 2 + 4 2 − r 2 c=AB=\sqrt{5^2-r^2}+\sqrt{4^2-r^2} and so on; the area is T = 1 2 r ( a + b + c ) = 19.8877 … T=\frac12 r(a+b+c)=19.8877\ldots giving the answer 198877 \boxed{198877} .

1 Helpful 1 Interesting 3 Brilliant 0 Confused

The way the r 4 r^4 terms cancel is suspiciously neat - it makes me think there's a quicker way to get to the cubic. Also, the cubic itself is a bit surprising - I think it means it wouldn't be possible to construct this with straight-edge and compass.

Oh, and in general, that cubic is 2 r 3 A I ⋅ B I ⋅ C I + r 2 [ 1 A I 2 + 1 B I 2 + 1 C I 2 ] = 1 \frac{2r^3}{AI\cdot BI \cdot CI}+r^2 \left[\frac{1}{AI^2}+\frac{1}{BI^2}+\frac{1}{CI^2}\right]=1

I think this always has a unique positive root, meaning for any A I , B I , C I AI,BI,CI there is a unique triangle, but I haven't looked into it too rigorously (Descartes' rule of signs seems enough, but I could be wrong).

Chris Lewis - 1 month, 2 weeks ago

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Re your last paragraph: Yes Descartes' rule of signs is the key here.

Another great solution by the way.

Pi Han Goh - 1 month, 1 week ago
Saya Suka
Apr 28, 2021

With an inradius x, I used 2 area formulae to get

x² = √[(9 – x²)(16 – x²)(25 – x²)] / [√(9 – x²) + √(16 – x²) + √(25 – x²)]

Then, using the positive real result of x ≈ 1.9002 to get the area of triangle
= x[√(9 – x²) + √(16 – x²) + √(25 – x²)]
= 19.8877955

Answer
= floor[ 10⁴ × 19.8877955 ]
= 198877

1 Helpful 0 Interesting 2 Brilliant 0 Confused

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