Square trapped in a triangle

let the side of the square be $x$ & the height of triangle $ADE$ be $h$ , therefore height of triangle $ABC$ is $h+x$ . Now, $[ADE] = 24, 1/2\times x\times h = 24$ , so $h = 48/x$ . Clearly, triangle $DGB$ & $EFC$ are congruent by RHS. Therefore, $BG=FC$ & $[DGB]=[EFC]$ . Now, $BC=48$ , $x+2\times BG=48$ , $BG=(48-x)/2$ .

Clearly, $2\times[DBG]+[ADE]+[DEFG]=[ABC]$ , which gives $2\times1/2\times((48-x)/2)(x)+24+x^2 =1/2(h+x)(48)$ , or that $2\times 1/2\times ((48-x)/2)(x)+24+x^2=1/2(48/x+x)(48)$ . By simplifying this we get, $x^3+48x-2304=0$ , which factorizes into $(x-12)(x^2+12x+192)=0$ solving this we get a real $x=12$ . therefore the side length of the square is 12.

[Latex edits - Calvin]

I worked this problem in 3 main steps.

[1] I divided the problem in half by dropping a perpendicular from A to the middle of BC.

[2] I came up with 2 equations and 2 unknowns; one equation using similar triangles, the other using the area of the triangle ADE.

[3] Combining the 2 equations and 2 unknowns, I solved for the value I needed.

Here are the details.

[1] I drew the isosceles triangle with A at the top, so that the left and right sides were equal. Since the triangle is isosceles, the sides AB and AC are equal and I could drop a perpendicular down from A to the middle of BC. This gives me a right triangle with base 24 and area of the upper triangle as 12. I labeled one side of the inscribed square with the variable $x$ . When the triangle was cut in half vertically, so was the square. So now I have a right triangle with an inscribed rectangle of height $x$ and width $x/2$ . On the perpendicular, I labeled the distance from A to the top of the inscribed rectangle as $z$ and the point on BC as F.

[2] For the first equation, I used the idea of similar triangles. The two triangles are the full triangle AFC and the small triangle at the top, 'sitting' on the inscribed rectangle. Comparing the bases and the vertical distances, I have $\frac{24}{x/2} = \frac{x+z}{z}$ <br /> The other equation comes from the area of the triangle given in the problem statement. This is the area of the small triangle, sitting on the inscribed rectangle.<br /> $A = \frac{1}{2} bh \to 12 = \frac{1}{2} \frac{x}{2}(z)$ <br />

[3] Doing algebra on the two equations, the variable $x$ is the value I want, so when I combine the equations, I eliminate $z$ . In the end, I have the equation<br /> $x^3 + 48x-48^2 = 0$ I chose to put that equation into my calculator to solve. There are 3 roots, only one of which is real, x=12.

Let a be the side length of the square DEFG . [ ABC ] = [ ADE ] * (\frac {48}{a})^2 = \frac {48^3}{2a^2}.

[ BDG ] + [ CEF ] =[ ADE ] * (\frac {48-a}{24})^2 = \frac {48^3}{2a^2} - \frac {48^2}{a} -24.

[ DEFG ] = [ ABC ] - [ ADE ] - ( [ BDG ] + [ CEF ] )

a^2 = \frac {48^3}{2a^2} - 24 - (\frac {48^3}{2a^2} - \frac {48^2}{a} -24)

a^2 = \frac {48^2}{a} - 48 ---(1)

a^3 + 48a = 48^2 ---(2) (I arranged so because 0 < a.)

By observing equation (1), since a and 48 are integers, \frac {48^2}{a} must also be an integer. Thus we can conclude * *a is a factor of 48 . By equation (2), 48^2 contains factors of 3 and since 48 is a multiple of 3, a^3 must at least be a multiple of 3. Furthermore, all the terms except a^3 in equation (2) are even. So a^3 must at least be a multiple of 2. Since a is an integer, * *a must be multiple of 6 . So possible value of a = 6, 12, 24 (48 is not possible, as the base length is already 48. Such square doesn't exist.) Substituting the value we get a = 12

Triangle ADE and triangle ABC are similar triangles as DE is parallel to GF [DEFG is a square] Let h be the height of the triangle ADE and x be the length of side of the square then height of traingle ABC is x+h Now, h/(x+h) = x/48 48h=x^2+hx h=x^2/(48-x) Now, Area of triangle ADE =1/2 x h 24=1/2 *x (x^2/(48-x)) 48=x^3/(48-x) 2304-48x =x^3 x^3+48x-2304=0 Solving this gives only one real solution and that is 12 Thus, x=12

Note: I'm not going to use LaTeX because I'm too lazy. Let s denote the side of the square, and let h denote the length of the altitude from A to BC. The area of triangle ADE is equal to half of DE times the height of triangle ADE, which is h-s. From this we get that s(h-s)/2=24. In triangle ABC, we can calculate the formula in two ways: 1/2 base altitude, or sum of the four divisions that the inscribed square makes. In the first method, we get that the area of the triangle is 24h. In the second method, BG and CF are both (48-s)/2. The area of the square is s^2, and the combined area of triangles BDG and CEF is s*(48-s)/2. From the results, we get that 24h=s(48-s)/2+s^2+24. When we calculated the are of ADE, we got that s(h-s)/2=24. So we alter the equation and make it h=(48+s^2)/s. Now we plug it in the first equation and move all of them to one side and make the coefficients simple, and it boils down to s^3+48s-48^2=0. Solving this, we get s=12.

Draw prependicular from A to AC meeting BC at M and DE at L. Triangles ALE and AMC are similar. DL=LE= 1/2 of side of square MC= 48/2=24 as ABC is isosceles triangle. If side of square is 2x, then LE=x and AL= 2*24/x= 48/x, AM= ax+48/x

now AL/AM = LE/MC = x/24 putting values and solving for x, we get x= 6 units side of square = 2x=2*6=12 units

Let $X$ and $Y$ be points on $DE$ and $BC$ respectively such that $AXY$ is perpendicular to $BC$ . Let $AX=h$ and $XY= s$ be the side of the square. Since $[ADE] = 24$ , we get $\frac {hs}{2} = 24$ or $hs = 48$ . Triangles $ADX$ and $ABY$ are similar, so $24 = BY = \frac {h+s}{h} DX = \frac {(h+s) s} {2h}$ . This gives $48 h = hs + s^2$ , so $s^2 + 48 = 48h$ . Multiplying throughout by $s$ , we get $s^3 + 48s = 48 hs = 48 \times 48$ . Since $s^3 + 48s$ is a strictly increasing function, this has a unique real root, which we can verify is $s = 12$ .

12