Where is the Vertex?

Geometry Level 5

A solid right circular cone is such that the angle between its axis and its surface is 2 0 20^{\circ} . A plane cuts through the cone at a certain angle with its axis, thus generating an elliptical base. The cone is placed on its elliptical base on a flat surface, such that the center of the ellipse coicides with the origin of an XYZ reference frame with the XY plane lying along the flat surface, and the positive Z-axis pointing upward. The XY axes orientation is such that the major axis of the elliptical base is along the X-axis and the minor axis is along the Y-axis.If the equation of the edge of the base is given by

x 2 25 + y 2 9 = 1 \frac{x^2}{25} + \frac{y^2}{9} = 1

What are the coordinates of the vertex? Assume that the vertex is located at ( a , b , c ) (a, b, c) where a 0 , b 0 , c 0 a \ge 0 , b \ge 0, c \ge 0 . Enter the value of a + b + c a+b+c as your answer.

Hint : Use the formulas in the solution of this problem .


The answer is 12.6578.

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1 solution

Hosam Hajjir
Aug 2, 2016

First, note that the major axis of the base lies in the same plane as the axis of the cone

and the normal to the cutting plane. Hence the axis of the cone lies in the

XZ plane. Therefore, the y-coordinate b = 0.

Next, note that, from the expressions for the semi-minor and semi-major axes

of the ellipse, we have that,

m / M = 1 sin 2 ( θ p ) sec 2 ( θ c ) m / M = \sqrt{1 - \sin^2(\theta_p) \sec^2(\theta_c)}

Hence,

3 5 = 1 sin 2 ( θ p ) sec 2 ( θ c ) \frac{3}{5} = \sqrt{1 - \sin^2(\theta_p) \sec^2(\theta_c)}

Since θ c = 2 0 \theta_c = 20^{\circ} , it is straight forward to calculate θ p \theta_p to be

θ p = 48.7 4 \theta_p = 48.74^{\circ}

Next, we need to find z 0 z_0 . Using the expression for the minor semi-axis (or equivalently the expression for the major semi-axis), we have

z 0 = m 1 sin 2 ( θ p ) sec 2 ( θ c ) / ( tan θ c cos θ p ) z_0 = m \sqrt{1 - \sin^2(\theta_p) \sec^2(\theta_c)} / ( \tan \theta_c \cos \theta_p )

From which,

z 0 = 3 ( 3 / 5 ) / ( tan 2 0 cos 48.7 4 ) = 7.499443211 z_0 = 3 (3/5) / (\tan 20^{\circ} \cos 48.74^{\circ} ) = 7.499443211

Note that the axis of the cone intesects the XY plane at

x 0 = z 0 sin θ p tan 2 θ c / ( 1 sin 2 ( θ p ) sec 2 ( θ c ) ) = 2.074598502 x_0 = z_0 \sin \theta_p \tan^2 \theta_c /(1 - \sin^2(\theta_p) \sec^2(\theta_c)) = 2.074598502

Now the vertex is now fully determined.

( a , b , c ) = ( 2.074598502 , 0 , 0 ) + 7.499443211 ( s i n ( 48.74 ) , 0 , c o s ( 48.74 ) ) = ( 7.7123 , 0 , 4.9455 ) (a,b,c) = (2.074598502 , 0, 0) + 7.499443211 (sin(48.74), 0, cos(48.74)) = (7.7123, 0, 4.9455 )

Therefore, the answer is 7.7123 + 0 + 4.9455 = 12.6578 7.7123 + 0 + 4.9455 = \boxed{12.6578}

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