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1. In Triangle ABC, the circumcenter and orthocenter are collinear with vertex A. Which of the following statements must be true?

(1) Triangle ABC must be an isosceles triangle.

(2) Triangle ABC must be an equilateral triangle.

(3) Triangle ABC must be a right triangle.

(4) Triangle ABC must be an isosceles right triangle.

Enter your answer as a comma-separated list. If there is no correct option, write "none".

2.

Let H be the orthocenter of the equilateral triangle ABC. We know the distance between the orthocenters of Triangle AHC and Triangle BHC is 12. What is the distance between the circumcenters of Triangle AHC and Triangle BHC?

3. Medians line AX and line BY of Triangle ABC intersect perpendicularly at point O. We know the lengths AX=12 and BC=4sqrt(13). Find the length of the third median CZ.

4.

The diagonals of a trapezoid are perpendicular and have lengths 3 and 4. Find the length of the median of the trapezoid.

Guest Dec 11, 2018

#1**+1 **

BX = 2sqrt(13) = sqrt (52) = CX

And since AX is a median, OX = (1/3)12 = 4

And since triangle OBX is right

OB = sqrt (BX^2 - OX^2) = sqrt (52 - 16) = sqrt (36) = 6

sin OXB = sin OXC = OB/BX = 6 / [ 2sqrt (13)] = 3 /sqrt (13)

Note that angle OXC is obtuse....so its cosine is negative

And cos OXC = - sqrt [ 1 - sin^2 OXC ] = -sqrt [ 1 - 9/13] = -sqrt (4 /13] = -2/sqrt (13)

Using the Law of Cosines

OC^2 = OX^2 + CX^2 - 2( OX * CX) cos OXC

OC^2 = 4^2 + 52 - 2 [4 * 2sqrt(13) ] (-2/sqrt (13)

OC^2 = 16 + 52 - 2 [ 4*2 ] (-2)

OC^2 = 16 + 52 + 32

OC^2 = 100

OC = 10

And this is 2/3 of CZ ....so....

(2/3)CZ = 10

CZ = 15

CPhill Dec 11, 2018