Day 17: Three Lengths in a Triangle

This is a very cool result! Somehow all the equations cancel to give a nice answer.

By Pythagoras' Theorem, $AB^2 + BC^2 = 60^2 \\ AB^2 + BE^2 = 52^2 \\ BD^2 + BC^2 = 39^2$ Now notice that adding the last two equations and taking away the first shows that: $(AB^2 + BE^2) + (BD^2 + BC^2) - (AB^2 + BC^2) = BD^2 + BE^2 = DE^2$ Therefore $DE^2 = 52^2 + 39 ^2 - 60^2 = 625$ And so $DE = \boxed{25}$ .

let $AD=b,EC=a,DB=x,BE=z$ by the Pythagorean theorem, we can say: $\begin{cases} x^2 +(z+a)^2 =39^2 \\ z^2 +(x+b)^2=52^2\\ (x+b)^2+(z+a)^2 =60^2 \end{cases}$ upon expansion, $\begin{cases} x^2 +z^2+2za+a^2 =39^2--(1) \\ x^2 +z^2 +2xb+b^2 =52^2--(2)\\ x^2+z^2 +2za+2xb+a^2+b^2 =60^2--(3) \end{cases}$ subtract 2 from 3 to get $2za+a^2 =896$ insert it in 1 to get $x^2 +z^2 +896= 39^2, x^2 +z^2=625$ by the Pythagorean theorem , $DE= \sqrt{DB^2+BE^2}=\sqrt{x^2 +y^2}=\sqrt{625}=\boxed{25}$

Let BE = x, BD = y, EC = a, AD = b

in triangle ABC (y + b)^2 + (x + a)^2 = 60^2 .......................................(1)

in triangle ABE (y + b)^2 + x^2 = 52^2 .................................................(2)

in triangle DBC (x + a)^2 + y^2 = 39^2 .................................................(3)

Adding (2), (3) and using (1) we get x^2 + y^2 = 52^2 + 39^2 - 60^2 = 625

in triangle DBE DE^2 = x^2 + y^2 = 625

Then DE = 25

Not a high level thinking question but a quite challenging one to get with it. Firstly looking at the ABC and applying pythagoras theorem in it [AC^2]=[AB^2 + BC^2]
3600=[AB^2 + BC^2] .........(1) Now, In triangle ABE, 2704=[AB^2 + BE^2] ...........(2) Now, In triangle DBC, 1521=[DB^2 + BC^2] ...........(3) Adding equations 2 and 3 : 4225=[AB^2 + BE^2 + DB^2 + BC^2 ] 4225=[3600 + BE^2 + DB^2] [From 1 ] [BE^2 + DB^2 ]=625 .......(4) But, [BE^2 + DB^2]=[DE^2} ............(5) [DE^2] = 625 DE=25 Here we got the required answer. ;)

Let BE = a, EC = b, BD = c, and AD = d. By Pythagorean theorem,

(1) (a+b)^2+(c+d)^2 = 60^2;

(2) a^2+(c+d)^2 = 52^2;

(3) (a+b)^2+c^2 = 39^2.

Performing the operation (2)+(3)-(1), we obtain a^2+c^2 = 52^2+39^2-60^2 = 625. Therefore DE = sqrt(a^2+c^2) = 25.