Mix and match cevians

We (can always) construct two cevians that concur with a median.

In a triangle, a cevian is a line segment from a vertex to the opposite side. A median is a special cevian that divides the opposite side in the ratio 1:1. Our aim in this post is to modify the ratios in which cevians divide the opposite sides, in such a way that concurrency is forced.

Ratios 1:2, 1:1, 1:2

In the diagram below, let cevian AL divide side BC such that BL:LC=1:2; let BM be a median, so that CM:MA=1:1; let cevian CN divide side BA in the ratio 1:2.

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We show that at the point of concurrency, median BM is bisected in the ratio 1:1, while cevians AL and CN are divided in ratios 3:1. (Compare this with the manner in which the centroid divides the three medians of a triangle.)

Using the given ratios, PROVE that AL,BM,CN are concurrent.

Just like proving that the three medians of a triangle are concurrent, this question belongs at the simplest level of simplicity. We have:

    \begin{equation*} \begin{split} \frac{BL}{LC}\times\frac{CM}{MA}\times\frac{AN}{NB}&=\frac{1}{2}\times\frac{1}{1}\times\frac{2}{1}\\ &=1 \end{split} \end{equation*}

By the converse of Ceva’s theorem, we conclude that AL,BM,CN are concurrent.

In the diagram below, the small letters p,q,r,s,t,u represent the areas of the triangles in which they are located. PROVE that u=3r, given that BN:NA=1:2, BL:LC=1:2, CM:MA=1:1.

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Let’s do this in four steps:

  • First, we have u=t. Consider \triangle AGM and \triangle CGM. They have the same altitudes, measured from their common vertex G. Since their base lengths CM and MA are also equal, their areas must be equal as well. This gives u=t.
  • Next we have s=2r and p=2q. Consider \triangle BGL and \triangle GLC. They have the same altitudes, measured from their common vertex G. Since BL:LC=1:2, the area of \triangle GLC is twice that of \triangle BGL, so s=2r. For the same reason we have p=2q, when \triangle AGN and \triangle BGN are compared.
    Our diagram simplifies to:

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  • We now show that q=r. Consider \triangle ABM and \triangle BMC. They have the same altitudes, measured from their common vertex B. Since CM:MA=1:1, these two triangles have the same areas, so u+2q+q=r+2r+u. This gives q=r.
  • Lastly, we show that u=3r. Consider \triangle ABL and \triangle ALC. They have the same altitudes, measured from their common vertex A. Since BL:LC=1:2, the area of \triangle ALC is twice the area of \triangle ABL. Thus, u+u+2r=2(2q+q+r), and so u=3q. Since q=r from the previous step, we get u=3r as desired.

Suppose that BL:LC=1:2, CM:MA=1:1, and BN:NA=1:2 in the diagram below:

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PROVE that at the point of concurrency, BM is bisected in the ratio 1:1, AL is divided in the ratio 3:1 and CN is divided in the ratio 3:1.

Let the point of concurrency be G and let the area of \triangle BGL be denoted by r. Then, in view of the previous example, the simplified area diagram, in terms of r, is:

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Consider \triangle ABG and \triangle AGM. They have the same areas, namely 3r. Since they also have the same altitudes (measured from their common vertex A), it must be the case that their base lengths are also equal. Thus BG=GM.

Now consider \triangle ACG and \triangle CGL. They have the same altitudes, measured from their common vertex C. The area of \triangle ACG is 6r, while the area of \triangle CGL is 2r. Thus their base lengths AG and GL must be in the ratio of their areas, so AG:GL=6r:2r=3:1.

Similarly we have CG:GN=3:1.

\triangle ABC has vertices at A(2,4), B(0,0), and C(6,0). Cevians AL,BM,CN divide sides BC,CA,BA in the ratios 1:2,1:1,1:2, respectively. Determine the coordinates of the point of concurrency.

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Note that cevian BM is a median. So it is bisected 1:1 at the point of concurrency, in view of the given ratios. This means that the midpoint of BM is the point of concurrency. Since M is the midpoint of CA, it is the point (4,2). Then we take the midpoint of (4,2) and B(0,0) to get (2,1) as the point of concurrency.

We can also find points L and N explicitly, and use the fact that cevians AL and CN are divided in the ratio 3:1 at the point of concurrency.

  • Since L divides BC in the ratio 1:2, it is the point \left(\frac{1\times 6+2\times 0}{1+2},\frac{1\times 0+ 2\times 0}{1+2}\right)=(2,0). Then the point of concurrency divides AL in the ratio 3:1, measured from vertex A. So it is the point \left(\frac{3\times 2+1\times 2}{3+1},\frac{3\times 0+ 1\times 4}{3+1}\right)=(2,1), as before.
  • Since N divides BA in the ratio 1:2, it is the point \left(\frac{2\times 0+1\times 2}{1+2},\frac{2\times 0+1\times 4}{1+2}\right)=\left(\frac{2}{3},\frac{4}{3}\right). The point of concurrency divides cevian CN in the ratio 3:1, measured from C. So it is the point \left(\frac{3\times 2/3+1\times 6}{3+1},\frac{3\times 4/3+1\times 0}{3+1}\right)=(2,1), as before.

So we obtain the same point (2,1), but the last two approaches involved more steps.

Ratios m:n, 1:1, m:n

This is a slight generalization of the previous examples.

In the diagram below, the small letters p,q,r,s,t,u represent the areas of the triangles in which they are located. PROVE that u=\frac{n}{2m}\left(\frac{m+n}{m}\right)r, given that BN:NA=m:n, BL:LC=m:n, CM:MA=1:1.

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As before, we do this in steps:

  • First we have u=t, because CM:MA=1:1 and both \triangle AGM and \triangle GMC have the same height.
  • Next we have s=\frac{n}{m}r. This follows because BL:LC=m:n and both \triangle GBL and \triangle GLC have the same altitudes, so their areas are proportional to their base lengths. Similarly, we have p=\frac{n}{m}q. At this stage, our simplified diagram is:

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  • Next we show that q=r. Consider \triangle ABM and \triangle BMC. Since CM:MA=1:1, these two triangles have the same areas, so r+\frac{n}{m}r+u=q+\frac{n}{m}q+u. That is, r+\frac{n}{m}r=q+\frac{n}{m}q. Re-arranging this gives (r-q)+\frac{n}{m}(r-q)=0, or (r-q)\left(1+\frac{n}{m}\right)=0. Since both m and n are positive (we’re dealing with internal division), we must have r-q=0, so r=q. Let’s adjust the diagram a bit:

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  • Lastly we show that u=\frac{n}{2m}\left(\frac{m+n}{m}\right)r. Consider \triangle ABL and \triangle ALC. They have the same altitudes, measured from their common vertex A. Since BL:LC=m:n, their areas follow the same ratio, meaning that \left(r+r+\frac{n}{m}r\right):\left(\frac{n}{m}r+u+u\right)=m:n.

        \begin{equation*} \begin{split} \frac{2r+(n/m)r}{(n/m)r+2u}&=\frac{m}{n}\\ n\left(2r+(n/m)r\right)&=m\left((n/m)r+2u\right)\\ \frac{n}{m}\left(2r+(n/m)r\right)&=(n/m)r+2u\\ \frac{n}{m}\left(2r+(n/m)r-r\right)&=2u\\ \frac{n}{m}\left(\frac{m+n}{m}\right)r&=2u \end{split} \end{equation*}

    Thus we get u=\frac{n}{2m}\left(\frac{m+n}{m}\right)r as desired.

Suppose that BL:LC=m:n, CM:MA=1:1, and BN:NA=m:n in the diagram below:

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PROVE that at the point of concurrency, BM is divided in the ratio 2m:n, AL is divided in the ratio (m+n):m, and CN is divided in the ratio (m+n):m.

Let the point of concurrency be G and let the area of \triangle BGL be r square units. Then, in terms of r, we have the following area diagram:

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Consider \triangle ABG and \triangle AGM. They have the same altitudes, measured from their common vertex A. So the ratio of their areas gives the ratio of their base lengths. In other words, BG:GM=\left(r+\frac{n}{m}r\right):\frac{n}{2m}\left(\frac{m+n}{m}\right)r.

    \begin{equation*} \begin{split} BG:GM&=\left(\frac{m+n}{m}\right)r:\frac{n}{2m}\left(\frac{m+n}{m}\right)r\\ &=1:\frac{n}{2m}\\ &=2m:n \end{split} \end{equation*}

To show that AG:GL=(m+n):m, consider \triangle AGC and \triangle GCL. They have the same altitude, measured from their common vertex C. So the ratio of their areas gives the ratio of their base lengths.

    \begin{equation*} \begin{split} AG:GL&=\left[\frac{n}{2m}\left(\frac{m+n}{m}\right)r+\frac{n}{2m}\left(\frac{m+n}{m}\right)r\right]:\frac{n}{m}r\\ &=\frac{n}{m}\left(\frac{m+n}{m}\right)r:\frac{n}{m}r\\ &=\frac{m+n}{m}:1\\ &=(m+n):m \end{split} \end{equation*}

Similarly, CG:GN=(m+n):m.

PROVE that the centroid of a triangle divides the three medians in the ratio 2:1.

In the case of three medians, we have m:n=1:1. So, using this in the previous example:

  • the median BM is divided in the ratio 2m:n=2(1):1=2:1;
  • the median AL is divided in the ratio (m+n):m=(1+1):1=2:1;
  • the median CN is divided in the ratio (m+n):m=(1+1):1=2:1.

In \triangle ABC, CN is a median. In what ratio should cevians AL and BM divide sides BC and CA so that the point of concurrency and the centroid trisect median CN?

Let G be the centroid, and let the point of concurrency be P. Then G divides median CN in ratio 2:1, measured from vertex C; that is, CG:GN=2:1. Since CN is to be trisected, we want CP:PG:GN=1:1:1, or CP:PN=1:2. From the previous example, the median CN is divided in the ratio 2m:n at the point of concurrency. Therefore 2m:n=1:2, or n=4m. So cevian AL should divide side BC in the ratio 1:4 and cevian BM should divide side CA in the ratio 4:1.

For any right triangle with legs parallel to the coordinate axes, PROVE that the slopes of its three medians form a geometric progression with common ratio -2.

For simplicity, fix the vertices at A(0,a), B(0,0), and C(c,0); a little modification will be needed for the general case.

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We now calculate the slopes of medians CN,BM,AL:

    \begin{equation*} \begin{split} \textrm{slope of median}~ CN&=-\frac{a}{2c}\\ \textrm{slope of median}~ BM&=\frac{a}{c}\\ \textrm{slope of median}~ AL&=-\frac{2a}{c} \end{split} \end{equation*}

These form a geometric progression with common ratio of -2.

(There’s something interesting here. In our first post of this year, we saw that if the slopes of the sides of a triangle form a geometric sequence with -2 as common ratio, then the triangle contains a vertical median and a horizontal median. Now, the above example says that if a triangle contains a vertical side and a horizontal side, then the slopes of its three medians form a geometric progression with a common ratio of -2. Did you notice something? See exercise 5 at the end.)

Give an example of a triangle in which one side, one median, and another cevian have slopes forming an arithmetic progression.

Consider \triangle ABC below, in which CN is a median, AL is a cevian:

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The slopes of AL,AC,CN are -5,-3,-1, respectively. They form an arithmetic progression with a common difference of 2.

(Notice that if a third cevian is drawn from vertex B and is concurrent with cevian AL and median CN, then it is an altitude. This is a good example of “mixing” and “matching” cevians as used in our post’s title. Mixing means that they are different; matching means that they are concurrent.)

Takeaway

The ratios we considered in this post (m:n,1:1,m:n) were for convenience. There are other ratios that can still ensure concurrency.

Tasks

  1. Let A(x_1,y_1),B(x_2,y_2),C(x_3,y_3) be the vertices of \triangle ABC. If there are cevians AL,BM,CN which divide sides BC,CA,BA in the ratios m:n,1:1,m:n, respectively:
    • PROVE that the point of concurrency has coordinates \left(\frac{mx_1+nx_2+mx_3}{2m+n},\frac{my_1+ny_2+my_3}{2m+n}\right);
    • Deduce that the centroid of \triangle ABC is the point \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).
  2. \triangle ABC has vertices at A(1,4),B(0,0),C(3,6). Assuming cevians AL,BM,CN divide sides BC,CA,BA in the ratios m:n,1:1,m:n, respectively, and they concur at the point \left(\frac{3}{2},\frac{15}{4}\right), determine m and n.
  3. \triangle ABC has vertices at A(2,4),B(0,0),C(6,0). Determine the lengths of cevians AL,BM,CN which divide sides BC,CA,BA in the ratios 1:3,3:1,1:1, respectively.
  4. Let \triangle ABC be a right triangle in the first quadrant, with vertices A(0,a), B(0,0), and C(c,0). For a> 1, PROVE that there are three cevians AL,BM,CN whose slopes form a geometric progression with common ratio -a.
    (Compare this with Example 9.)
  5. Let A(x_1,y_1),B(x_2,y_2),C(x_3,y_3) be the vertices of \triangle ABC. Let the slopes of medians through A,B,C be denoted by m_A,m_B,m_C, respectively. Suppose that m_C,m_B,m_A (in that order) form a geometric progression with a common ratio of -2. PROVE that:
    • y_2=y_3;
    • x_1=x_2.
      (In view of Example 9, conclude that the slopes of the three medians of a triangle form a geometric progression with common ratio -2 if, and only if, the triangle is a right triangle with legs parallel to the coordinate axes.)
  6. (Opposite slopes) The right triangle ABC with vertices at A(0,3), B(0,0), C(4,0) has very nice properties:

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    • Find an internal point L on BC such that the slope of cevian BX is the negative of the slope of cevian AL, and PROVE that these two cevians intersect at \left(\frac{2}{3},\frac{3}{2}\right);
    • Find an internal point N on AB such that the slope of cevian BZ is the negative of the slope of cevian CN, and PROVE that these two cevians intersect at \left(2,\frac{1}{2}\right);
    • If a cevian from B and a cevian from C have opposite slopes, is their point of intersection always of the form \left(a,\frac{1}{a}\right) for some a?
    • For any point R between A and M, PROVE that there is a point S between B and C such that the slopes of cevians BR and AS are negatives of each other;
    • For any point U between M and C, PROVE that there is a point V between A and B such that the slopes of cevians BU and CV are negatives of each other.
  7. Consider the right triangle ABC with vertices at A(0,a), B(0,0), C(c,0):

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    An arbitrary point X on AC can be given by the coordinates \left(k,-\frac{a}{c}k+a\right), where 0< k< c. PROVE that:

    • \frac{ck}{c-k}< c if and only if k<\frac{c}{2};
    • \frac{a}{k}(c-k)< a if and only if k> \frac{c}{2};
    • for k<\frac{c}{2} and L:=\left(\frac{ck}{c-k},0\right), cevians AL and BX have opposite slopes;
    • for k> \frac{c}{2} and V:=\left(0,\frac{a}{k}(c-k)\right), cevians CV and BX have opposite slopes;
    • a cevian from B and a cevian from C intersect at y=\frac{a}{2} if and only if they have opposite slopes (and k<\frac{c}{2});
    • a cevian from B and a cevian from C intersect at x=\frac{c}{2} if and only if they have opposite slopes (and k>\frac{c}{2}).
  8. Let A(x_1,y_1),B(x_2,y_2),C(x_3,y_3) be the vertices of \triangle ABC. Suppose that the slopes of sides AB and AC are negatives of each other, and that the slopes of medians through B and C are also negatives of each other. PROVE that:
    • 2x_1=x_2+x_3 (so x_1,x_2,x_3 form an arithmetic sequence)
    • AB=AC (so \triangle ABC is isosceles)
    • the median through vertex A is vertical.
  9. PROVE that if the slopes of the equal sides of an isosceles triangle are negatives of each other, then the x-coordinates form an arithmetic progression.
  10. For an integer r> 1, let A(1,r^2),B(0,0),C(1+r,r+r^2) be the vertices of \triangle ABC. PROVE that a cevian from vertex A divides side BC in the ratio (r^2-r+1):r if and only if it is an altitude.