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Serga [27]
2 years ago
11

Wanda is trying to locate the Fermat point $P$ of $\triangle ABC$, where $A$ is at the origin, $B$ is at $(8,-1)$, and $C$ is at

$(5,4)$ (the Fermat point is the point such that the sum of its distances from the vertices of a triangle is minimized). She guesses that the point is at $P = (4,2)$, and computes the sum of the distances from $P$ to the vertices of $\triangle ABC$. If she obtains $m + n\sqrt{5}$, where $m$ and $n$ are integers, what is $m + n$?
Mathematics
1 answer:
lidiya [134]2 years ago
6 0

Answer: 8

Step-by-step explanation:

The distance formula is used to find the distances to each of the points:

 d = √((x2-x1)² +(y2-y1)²)

 d = √((4-0)² +(2-0)²) +√((4-8)² +(2-(-1))²) +√((4-5)² +(2-4)²)

 d = √20 +√25 +√5 = 2√5 +5 +√5

 d = 5 +3√5

Comparing to the form

 d = m +n√5

we see that m=5, n=3. The the sum m+n = 5+3 = 8.

(sorry if im wrong)

:(

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============================================================

Explanation:

I'm assuming that segments AD and CD are tangents to the circle.

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Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.

Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.

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---------------------------

Here's what we have so far for quadrilateral DAEC

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