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December 30, 2015

Novak Djokovic against younger players

Filed under: sport — ckrao @ 9:11 pm

Novak Djokovic had a stellar tennis record in 2015, winning three out of the four grand slams (runner-up in the other), the ATP World Tour finals for the fifth time, a record six Masters titles and reached 15 consecutive finals (winning 11 of them) in attaining an 82-6 record (31 wins against top 10 players!!) and US$21.6m for the year.

I noticed that all his losses for the year were to players older than him and I looked up his record against younger players. Here is a list of the players younger than Djokovic who have beaten him in his entire ATP career. The month refers to when the associated tournament started.

  • Ernests Gulbis (Jan 2009)
  • Filip Krajinovic (May 2010 – Djokovic retired after losing the first set)
  • Juan Martin del Potro (Sep 2011, Jul 2012 and Mar 2013)
  • Kei Nishikori (Oct 2011 and Aug 2014)
  • Sam Querry (Oct 2012)
  • Grigor Dimitrov (May 2013)

That’s right, only six players have had this honour for a total of nine losses. After his loss to Dimitrov at the Madrid Masters he has won 71 of his last 72 matches against younger players, losing only to Nishikori in the 2014 US Open semi-finals. He is currently on a 40-match winning streak and to the end of 2015 his record is 130-9 against younger players (career: 686-146). As a comparison here is the record of Murray, Nadal and Federer against younger players not named Novak Djokovic, Andy Murray or Rafael Nadal.

  • Murray: 118-19
  • Nadal: 177-17 (including 41 in a row from Nov 2009 to the end of 2011)
  • Federer: 517-61 (the higher numbers show the clear generation gap)



Tennis Abstract: ATP and WTA Match Results, Splits, and Analysis

December 23, 2015

Areas of sections of a triangle from distances to its sides

Filed under: mathematics — ckrao @ 12:35 pm

If a point P is in the interior of triangle ABC distance x, y and z from the sides, what is the ratio of the area of quadrilateral BXPZ to that of ABC?


One way of determining this is to draw parallels to the sides of the triangles through P. Let X_1 and X_2 be where these parallels meet side BC as shown below.


Let the sides of the triangles have lengths a, b, c with corresponding altitudes h_a, h_b, h_c.

Then as \triangle PX_1 X_2 and \triangle ACB are similar,

\begin{aligned}|PX_1X| &= |PX_1X_2| \frac{X_1X}{X_1X_2}\\ &= |PX_1X_2| \frac{b\cos C}{a}\\ &= |PX_1X_2| \frac{b (a^2 + b^2 - c^2)}{2a^2b} \quad \text{ (cosine rule)}\\ &= \left(\frac{x}{h_a} \right)^2|ABC|\frac{(a^2 + b^2 - c^2)}{2a^2}\\ &= |ABC|\left(\frac{ax}{ax+by+cz}\right)^2\frac{(a^2 + b^2 - c^2)}{2a^2}\\&= \frac{|ABC|x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2},\quad\quad (1) \end{aligned}

where the second last line follows from twice the area of |ABC| being ah_a = ax + by + cz.


\displaystyle |PY_1Z| = \frac{|ABC|z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}.\quad \quad (2)


\begin{aligned}|X_1Y_1B| &= \left(\frac{h_b-y}{h_b}\right)^2|ABC|\\ &= \left(1-\frac{by}{bh_b}\right)^2 |ABC|\\ &= \left(1-\frac{by}{2|ABC|}\right)^2|ABC|\\ &= \left(1-\frac{by}{ax+by+cz}\right)^2|ABC|\\ &= \left(\frac{ax +cz}{ax+by+cz}\right)^2|ABC|. \quad\quad(3)\end{aligned}

Combining (1), (2) and (3), we obtain our desired answer as

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\&= \left(\frac{ax +cz}{ax+by+cz}\right)^2-\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\&=\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\ &= \frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\&= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\quad\quad(4)\end{aligned}

Similar formulas can be found for quadrilaterals XPYC and YPZA by permuting variables. Note that if P is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables x, y, z.

Note that (4) may also be written as

\displaystyle \frac{|BXPZ|}{|ABC|} = \frac{ac(2xz + (x^2 + z^2)\cos B)}{(ax+by+cz)^2}.\quad\quad(5)

Special cases

1) If \triangle ABC is equilateral, a=b=c and from (4) we obtain

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(6)\end{aligned}

2) If P is at the incentre of \triangle ABC, then x = y = z = r (the inradius) and from (4) we have

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(7)\end{aligned}

3) If \triangle P is right-angled at B, then quadrilateral BXPZ is a rectangle with area xz and \triangle ABC has area ac/2 and from (5),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{2acxz )}{(ax+by+cz)^2}\\ &= \frac{2acxz )}{(ac)^2}\\ &= \frac{2xz}{ac}.\quad \quad (8)\end{aligned}

as expected.

4) If a=c and x=z (symmetric isosceles triangle case) then from (4),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2x^2 + 2x^2(2a^2-b^2)}{2(2ax+by)^2}\\ &= \frac{x^2(4a^2 -b^2)}{(2ax+by)^2}.\quad\quad(9)\end{aligned}

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