the problem will blow your mind

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[@Irtsak]

A different finish in Math Booster's solution . Math booster estimated (a,b)=(6,10) or (a,b)=(15,17) In Greece , in high school textbooks , there is a formula : E=τ⋅ρ Ε= area of the triangle , τ=semi perimeter of the triangle and ρ = the radius of the inscribed circle in the triangle. E=τ⋅ρ=>ρ=E/τ If a=6 , b=10 and the other side =8 , the semi perimeter is τ=12. Area (ABC)=24 . So ρ=(ABC)/τ=24/12=2 If a=15 , b=17 and the other side =8 , the semi perimeter is τ=20. Area (ABC)=60 . So ρ=(ABC)/τ=60/20=3 **** We use E for area cause the Greek word for area is (Emvado)

[@prime423]

Most Mathletes would recognize that the triangle sides are clearly 6,8 and 10.3-4-5.Since the sides are not given and are integral ,this fits the bill. No manipulation required!!

[@michaeldoerr5810]

I actually agree with you. I think that is an example of a geometry problem that should be taught as easier than it looks. It is just a combination and a Pythagotrean triplet and a relevant construction. Also, does that make me a mathlete?

[@hongningsuen1348]

The second part of solution can be done using the formula: area of triangle = radius of triangle's incircle x semiperimeter of triangle For the 2 solutions of this problem: area of triangle = 24 or 60 semiperimeter of triangle = 12 or 20 radius of incircle = 2 or 3.

[@jimlocke9320]

Wrapping up at 10:40 using the tangent double angle formula: Construct AO. ΔAOE and ΔAOM are congruent,

[@prime423]

rs=K.Where r is the radius of the inscribed circle and s equals semiperimeter.

[@jonpress6773]

I started by drawing the radii to the three points of tangency between the triangle and the circle, and then the line segments from the center of the circle to each of the three vertices of the triangle. This breaks the triangle into 6 smaller right triangles. Then I planned to add the areas of these 6 triangle and compare this sum to the area of the entire triangle, (AB)*(BC)/2. We have all the lengths we need to do this, except the length from C to the tangency point on BC. I called that x. The resulting area comparison was (8-R)R + Rx + R^2 = 8(R+X)/2. We need another relationship between R and x, which comes from the Pythagorean Theorem applied to triangle ABC: We have expressions for all the sides: AB = 8, BC = R + x, and after a little manipulation, AC = 8 - R + x. So 64 + (R+x)^2 = (8-R+x)^2. I solved for x in the original area comparison, to get x = 4R/(4-R) and substituted that into 64 + (R+x)^2 = (8-R+x)^2. I figured I would get some equation for R that I could solve -- hopefully easily. It looked messy, with 3rd and 4th powers of R showing up. Hopefully, they would cancel, or at least the odd powers would cancel so that I could get a quadratic equation in R^2. But when all the cancelling was done, there was nothing left! 64 + (R+x)^2 was IDENTICALLY equal to (8-R+x)^2. I realized that the driving force behind this being an identity was the Pythagorean Theorem itself, and that maybe this construction could be used in reverse to prove the Pythagorean Theorem. That turned out to be the case, and I stumbled on what I hoped against hope was actually a new proof! But no, it's been done. But definitely unexpected.

[@klatis84]

good one

[@soli9mana-soli4953]

There are 2 pythagorean triplets (8 , 6 , 10) and (8 , 15 , 17)...

[@giuseppemalaguti435]

Vista così, sicuramente manca un dato...ah!!ok,integers...può essere AC=17...BC=15....R=A/p=8*15/20=6

[@harrymatabal8448]

Reading the comments I realized that there are a whole lot of intelligent mathematician around. Congratulations. But the author thinks we are stupid. She thinks we don't know the theorem of Pythagoras. Time wasting.