This is a paragraph.

A sample problem from social media I

Examine the following problem from a Facebook group:

Solve the above problem. (See below for enlarged image.)

Sample solution here .

Takeaway

Let ABC be a non-right triangle with side-lengths a,b,c, and let T_b be as defined in today’s post. Then the following statements are equivalent:

  1. OT_b=\frac{2R^2}{a}
  2. (b^2-a^2)^2=(ac)^2+(cb)^2.

Task

  • Let ABC be a triangle with vertices at A(-6,0), B(0,0), and C(2,4). Let T_b be as defined in today’s post.
    1. Find the coordinates of T_b
    2. Verify that AT_b=2R=CT_b, where R is the circumradius.

An example of harmonic division II

Let ABC be a right triangle in which \angle C=90^{\circ}, and let F_c be the foot of the altitude from C. The segment CF_c contains two special points: the symmedian point (S), and the Kosnita point (K). We show an example where S and F_c divide CK harmonically.

Given \triangle ABC with vertices located at A(0,6), B(8,0), and C(0,0), find the symmedian point.

Observe that the given triangle is right-angled at C. Further, in a right triangle, the symmedian point is the midpoint of the right-angled vertex and the foot of the altitude from the right-angled vertex. In a previous post, we found the foot of the altitude from C to be F_c\left(\frac{72}{25},\frac{96}{25}\right). Thus, the symmedian point in this case is S\left(\frac{36}{25},\frac{48}{25}\right).

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Given \triangle ABC with vertices located at A(0,6), B(8,0), and C(0,0), verify that S and F_c divide CK harmonically, where K is the Kosnita point, S is the symmedian point, and F_c is the foot of the altitude from C.

We have C(0,0), S\left(\frac{36}{25},\frac{48}{25}\right), K\left(\frac{48}{25},\frac{64}{25}\right), and F_c\left(\frac{72}{25},\frac{96}{25}\right).

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Explicit calculation gives

    \[CS=\frac{12}{5}, SK=\frac{4}{5}, CF_c=\frac{24}{5},KF_c=\frac{8}{5}\]

Thus S and F_c divide CK harmonically, since:

    \[\frac{CS}{SK}=\frac{3}{1}=\frac{CF_c}{KF_c}.\]

Takeaway

In any right triangle, the following statements are equivalent:

  1. the right triangle is isosceles
  2. the Kosnita point coincides with the centroid.

Task

  • (Equivalent statements) Let ABC be a non-right triangle. PROVE that the three statements below are equivalent:
    1. 2\cos A\cos B+\cos C=0
    2. 2\sin A\sin B-\cos C=0
    3. \tan A+\tan B+2\tan C=0
  • (Equal steps) Suppose that triangle ABC satisfies any of the three equivalent statements above. In such a triangle, let S_c be the foot of the symmedian from C, and let F_a,F_b be the feet of the altitudes from A,B in that order. PROVE that F_aS_c=F_bS_c=\frac{abc}{a^2+b^2}.