Allrighty. It's pretty long, and in 3 different major sections.
Section one: Proving that the angle between gray and *pink* (I misspoke in a previous post) is 90 degrees.
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The major formula that I used for this problem is that the area of a triangle is equal to .5absin(C). The other formula I used is sin(A+B)=sin(A)cos(B)+cos(A)sin(B).
Legend:
BD=y
AD=x
BC=w
AE=d
<ABE=A
<EBC=B
From the first formula mentioned, we get the area of the whole triangle is .5(x+y)(w)(sin(A+B)).
The whole triangle is also the sum of the areas of triangles ABE and BEC.
Those areas (by the same formula) are .5(x+y)(d)(sin(A)) and .5(w)(d)(sin(B)).
Thus, .5(x+y)(w)(sin(A+B))=.5(x+y)(d)(sin(A))+.5(w)(d)(sin(B)).
The .5's divide out of the equation. I will keep sin(A+B) on the left.
sin(A+B)=(d/w)sin(A)+(d/(x+y))sin(B).
From the second formula...
sin(A)cos(B)+cos(A)sin(B)=(d/w)sin(A)+(d/(x+y))sin(B).
This means that cos(B)=(d/w) and cos(A)=(d/(x+y)). Those are exact matches to their respective "adjacent" and "hypotenuse" sides, in each of the smaller triangles.
That only happens in a right triangle. Therefore, <E is right.
A similar argument shows that <D is right.
Section two: Marking up congruent parts of the triangle.
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×We now have shown that the gray triangles have one angle the same. They also have a pair of vertical angles. This means they are similar, as they have two angles the same.
Since they have the same area as well, they are congruent.
From CPCTC, we get:
DF=FE
DB=EC
BF=FC
<DBF=<ECF
Since BF=CF, triangle BCF is isosceles, making <FBC=<FCB. This makes the entire triangle isosceles. Thus AB=AC, and AD=AE.
BE and CD are altitudes, and they meet at F. The three altitudes of a triangle meet at a point (I forget the name--I think it's the orthocenter.) This means that AF is an altitude. Extend AF to meet BC, and we get another right triangle. Call that meeting point G.
Section three: Solving for the area.
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×In this section, I will reassign some variables.
Legend:
FE=w
FC=x
EC=y
AE=z
Again, we have many right triangles here.
Since ADFE's area is 25, the area of triangle AFE is 12.5 (half of a kite).
Also, area of FGC (G is defined in section 2) is 3.5, as it's half of an isosceles triangle (cut by the altitude.)
This means that
.5wz=12.5
.5wy=X (Area of a gray triangle)
Now, look at the medium triangles (blue and gray, pink and gray.)
.5(w+x)(y)=7+X
.5(w+x)(z)=25+X
Working with the first...
.5wy+.5xy=7+X
X+.5xy=7+X (substitution)
.5xy=7
Working with the second...
.5wz+.5xz=25+X
12.5+.5xz=25+X
.5xz=12.5+X
So we now have
.5wz=12.5
.5wy=X
.5xy=7
.5xz=12.5+X
For simplicity, let's multiply everything by 2.
wz=25
wy=2X
xy=14
xz=25+2X
From these, we can solve for X. (Major use of substitution)
y=2X/w
x=14w/2X
25+2X=14wz/2X=350/2X
50X+4X^2=350
4X^2+50X-350=0.
Solving this quadratic equation yields that X=5 or X=-18. X is an area, so must be positive. X=5.
5+5+7+25=42.
There's probably a MUCH shorter way to solve it that I didn't see, though.
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First Delver! (I was the first non-tester/dev
to conquer TCB.)
d
/dy
