Triangle Brain Teaser

Geometry Level 3

$\triangle ABC$ (blue) has side length 5, 6, and 7. $\triangle A' B' C'$ (red) is drawn inside it with its sides parallel to the sides of $\triangle ABC$ , and 1 unit away from them. What is the length of $A' B'$ ? Round your answer to 3 decimal places.

The answer is 1.938.

2 solutions

Maria Kozlowska
Sep 4, 2017

Inradius of $\triangle ABC$ is $r=\dfrac{2 \sqrt{6}}{3}$ . Inradius of $\triangle A'B'C'$ is $r'=\dfrac{2 \sqrt{6}}{3}-1$ .

$A'B' = AB \dfrac{r'}{r} = \dfrac{-5 \sqrt{6} + 20}{4} \approx 1.938$ .

Exactly what I had in mind. Thanks for posting the solution.

Hosam Hajjir - 3 years, 9 months ago

My approach was the following : I extended the sides of triangle A'B'C' till they met BC and then attempted to find B'C' as BC - 1/tan(B) - 1/tan(C) - 2 (after using cosine rule to get cos(B) & cos(C) ) and then calculated A'B' as B'C' * 5 / BC but ai get a different answer , around 2.381

Sundar R - 3 years, 9 months ago

B'C' = BC - 1/sin(B) - 1/sin(C) - 1/tan(B) - 1/tan(C)

Hosam Hajjir - 3 years, 9 months ago

Thanks ! I realize the mistake. I suppose the 1/sin(A) - 1/sin(B) replaces the 2.

Sundar R - 3 years, 9 months ago

Rab Gani
Dec 9, 2017

Connect point AA’, BB’, CC’.Let A’B’, A’C’, B’C’ are x,y,z,respectively .The Area of ΔA’B’C’ = 6√6 – ½ (x+y+z +18). By similarity, the area ratio : [ΔA’B’C’]/ [ΔABC] = [6√6 – ½ (x+y+z ]/ 6√6. = (x/5)^2, By similarity (x+y+z)/18 = x/5, [6√6 – 9(x/5)]/ 6√6. = (x/5)^2, So we have quadratic eqs. in (x/5) :
(x/5)^2+0.61237 (x/5) – 0.3876 = 0, then x=1.938

Triangle Brain Teaser

The answer is 1.938.

2 solutions

0 pending reports