Given a triangle with vertices $P$, $Q$, and $R$, where $\angle PQR = 90^\circ$, $\angle QRP = 30^\circ$, and the length of side $PQ = 8$ units. An ellipse is inscribed in this triangle, touching side $PQ$ at its midpoint and each of the other two sides at exactly one point. The major axis of the ellipse lies along the angle bisector of $\angle QRP$. Calculate the length of the minor axis of the ellipse."
#Ellipse inscribed in a 30-60-90 triangle
cedar sinew
warm wagonBOT
ZiXO
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strange hearthBOT
Given a triangle with vertices $P$, $Q$, and $R$, where $\\angle PQR = 90^\\circ...
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warm relic
Given a triangle with vertices $P$, $Q$, and $R$, where $\\angle PQR = 90^\\circ...
you remember the formula for an ellipse, right?
(im going to propose a very odd solution if you do)
(because you can coord bash if you know)
cedar sinew
i think it was $ (A = \pi \cdot {a} \cdot{b})$
warm relic
i think it was $ (A = \pi \cdot {a} \cdot{b})$
no, not the area formula
actually, i dont think we need it for this.
we can find the major axis very easily
cedar sinew
using the triangle sides lengths ?
warm relic
using the triangle sides lengths ?
uh, kind of
let us denote point Q to be the origin
thus point P is (0,8) and point R is (8sqrt(3),0)
to find the slope of the line, we can simply find the tangent of the angle it makes with the x-axis
we can clearly see this occurs at point R
since the major axis is an angle bisector
we have the equation for the major axis being y-0=tan(15)(x-8)
simplifying we get y=tan(15)x-8tan(15)
do you see where im going?
warm relic
yes of course, Thanks man i was about to give up on this
lol (never give up on math problem)
yw!
cedar sinew
lol (never give up on math problem)
True, every time i solve a new problem for me it gives me a new perspective to keep in mind for new problems
+close