# Location of a triangular from several of its components

I located this inquiry a while earlier on a SAT technique test or something, can not fairly bear in mind. So offered an intense triangular $ABC$ with $P$ a factor inside it and also $AP$, $BP$, and also $CP$ conference the contrary sides at $D$, $E$, and also $F$ specifically:

Just how can you locate the location of triangular $ABC$ offered the locations of triangles $x$, $y$, and also $z$?

**Method 1 ** : We can execute an affine transformation to make ADC any kind of triangular we desire, while maintaining colinearality and also the proportions of locations. Specifically, this permits us to assign specific works with to A, D and also C. We can after that make use of the proportion of x and also y to locate E and also the proportion of z and also x+y+z to locate P. We can after that make use of these works with to locate B. As soon as we understand B, we understand the proportion of BC to DC and also therefore the proportion of the location of the entire triangular to the well-known locations.

**Method 2 ** : Let Q be the location of the entire triangular

```
AE:EC=x:y=|AEB|:|BEC|
|APB|=xQ/(x+y)-x
|BPD|=yQ/(x+y)-y-z
Now AP:PD=x+y:z
z(xQ/(x+y)-x)=(x+y)(yQ/(x+y)-y-z)
Q(zx/(x+y)-y)=xz-xy-y^2-xz-yz=-xy-y^2-xz
Q(zx-xy-y^2)/(x+y)=(-xy-y^2-xz)
Q=(-xy-y^2-xz)*(x+y)/(zx-xy-y^2)
```

This was not most likely to have actually been an SAT technique trouble, though it is a regular competition trouble.

$AE:EC = x:y$ (given that those 2 triangulars have the very same elevation to $AC$, the proportions of their locations is the proportions of their bases relative to that elevation) and also $AP:PD = (x+y):z$ (very same suggestion as $AE:EC$). Recognizing these 2 proportions, use the strategy of mass points, placing masses $zy$ at $A, zx$ at $C$ (offers the proportion $x:y$ for $AE:EC$), and also $y(x+y)$ at $D$ (offers $(x+y):z$ for $AP:PD$). This causes a mass of $y(x+y)-zx$ at $B$, so the proportion $BD:DC = zx:(y(x+y)-zx).$ This have to additionally be the proportion of the locations of △ ABD to △ ADC (usual elevation once more), so (location of △ ABD) :$(x+y+z) = zx:(y(x+y)-zx)$. Addressing from there refers slamming out the algebra.

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