Chaitanya's Random Pages

July 27, 2013

Wide range in maximum temperatures within a month in Melbourne

Filed under: climate and weather,geography — ckrao @ 5:02 am

This month Melbourne recorded its highest July recorded maximum temperature of 23.3°C. Then just two days later the maximum was just 9.7°C. This temperature difference seemed highly unusual to me and indeed it was. The difference of 13.6°C is easily the highest recorded for the month of July in over 150 years of records (the previous record was 12° in 1975 when it was 11.1°C on the 3rd and 23.1°C on the 30th of the month). The graph below shows the distribution of the difference between highest maximum and lowest maximum temperatures for each month (data from the Australian Bureau of Meteorology website). In a box and whisker plot the thick horizontal lines represent the medians and the box boundaries are the 25th and 75th percentiles. Outliers are shown for data more than 1.5 times the inter-quartile range away from the 25th or 75th percentiles. We see that in Melbourne much larger maximum temperature fluctuations are expected in the warmer months. July has a median range of just 7.7°C between its highest and lowest maximum temperatures.


Here are a few other outliers indicated on the graph when there were large extremes in maximum temperatures during the month.

  • Dec 1867: 10.4°C (12th – coldest Dec maximum on record), 40.3°C (19th)
  • May 1905: 28.7°C (9th – warmest May day on record), 11.4°C (11th)
  • Nov 1911: 12.2°C (1st), 40.7°C (30th)
  • Oct 1922: 35.8°C (22nd), 9.0°C (29th – coldest Oct maximum on record)
  • Dec 1924: 40.1°C (12th), 11.5°C (26th)

It’s interesting that it has been so long since we have had such an anomaly in maximum temperatures within a month!


July 25, 2013

Solving a tricky geometry problem using cross ratios

Filed under: mathematics — ckrao @ 12:17 pm

I recently had fun playing around with the following problem, based on the last question of the 2003 AIME I.

In triangle ABC let M be the midpoint of CA, and let D be the point on CA such that BD bisects angle ABC. Let F be the point on BC such that DF and BD are perpendicular. If DF meets BM at E, find the ratio DE/EF in terms of the side lengths of ABC.

AIME_I_2003_Q15The interested reader might like to have an attempt at this problem before reading further. 🙂

Initially I solved this using vector geometry applying facts such as \vec{BM} = (\vec{BA} + \vec{BC})/2, \vec{BD} = (|BC|\vec{BA} + |AB|\vec{BC})/(|BC| + |AB|) and \vec{DB}. \vec{DF} = 0 (see below). After seeing it led to a relatively simple answer I started looking for a more elegant approach and after seeing this page that shows methods using mass point geometry and applications of Menelaus’ theorem, came up with the following nice solution.

Extend FD to meet BA in G as shown below. As BD is an angle bisector perpendicular to GF, F reflects in line BD to G so D is the midpoint of GF.


Also the lines ADMC and GDEF are both perspective from the point B, so the cross ratios (AM/MC)/(AD/DC) and (GE/EF)/(GD/DF) are equal (indeed, by using the triangle area formula \frac{1}{2}ab\sin C these ratios can be shown to be equal to (\sin \angle ABM/ \sin \angle MBC)/(\sin \angle ABD / \sin \angle DBC)). Since M and D are midpoints of AC and GF respectively, AM/MC = GD/DF = 1, so we are left with

\displaystyle \frac{GE}{EF} = \frac{DC}{AD}.

By the angle bisector theorem, this ratio is equal to a/c, where a = BC, c = AB. We finally have

\displaystyle \frac{DE}{EF} = \frac{GE-EF}{2EF} = \frac{GE/EF - 1}{2} = \frac{a-c}{2c},

as required. It’s interesting that the answer has no dependence on the length of side AC.

Here is the original way I had solved the problem. Let B be the origin and define vectors \boldsymbol{a} = \vec{BA} and \boldsymbol{c} = \vec{BA} of length c and a respectively. Then as BD is the angle bisector,  we can write \vec{BD} = \lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}, where \lambda_1 = a/(a+c) and \lambda_2 = c/(a+c).

Also let \vec{BF} = k \boldsymbol{c} where k = BF/BC. We find k by using the fact that DF \perp DB:

\displaystyle \left(kc - (\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})\right).(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}) = 0.

From this,

\displaystyle k = \frac{(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}).(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}{\boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})} = \frac{BD^2}{\boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}.

Since E is on DF, for some \mu between 0 and 1,

\displaystyle \vec{BM} = \mu \vec{BD} + (1-\mu) \vec{BF} = \mu \lambda_1 \boldsymbol{a} + (\mu \lambda_2 + (1-\mu) k )\boldsymbol{c}.

Since E is also on the median BM, it and have equal components of a and c (being a median).

Hence, \mu \lambda_1 = \mu \lambda_2 + (1-\mu) k from which

\displaystyle \frac{\lambda_1 - \lambda_2}{k} = \frac{1-\mu}{\mu} = \frac{(\lambda_1 - \lambda_2) \boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}{BD^2}. \quad \quad (1)

We wish to find the ratio DE/EF = (1-\mu)/\mu, which is as above.

Here is one easy way to find k = BF/BC that I found only afterwards. If P is the midpoint of BF then since BDF is a right-angled triangle, BP = PD = PF and so \angle PDB = \angle DPE = \angle DBA, from which AB \parallel DP.


Then \triangle BCA and \triangle PCD are similar, from which

\displaystyle \frac{BP}{BC} = \frac{AD}{AC} = \frac{c}{a+c}.

Hence k = BF/BC = \frac{2c}{a+c} and so

\displaystyle \frac{DE}{EF} = \frac{\lambda_1 - \lambda_2}{k} = \frac{(a-c)}{(a+c)} \frac{(a+c)}{2c} = \frac{a-c}{2c}.

The original way I had found k was without the benefit of this construction and by pure computation instead. Referring back to (1), we have \boldsymbol{c}.\boldsymbol{c} = a^2 and by the cosine rule, \boldsymbol{c}.\boldsymbol{a} = (a^2 + c^2 - b^2)/2. Recalling that \lambda_1 = a/(a+c) and \lambda_2 = c/(a+c),

\begin{aligned} \boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}) &= \lambda_1 \boldsymbol{c}.\boldsymbol{a} + \lambda_2 \boldsymbol{c}.\boldsymbol{c}\\ &= \frac{a(a^2 + c^2 - b^2) + 2ca^2}{2(a+c)} \\ &= \frac{a(a+b+c)(a-b+c)}{2(a+c)}. \quad \quad (2) \end{aligned}

Secondly the denominator of (1) is

\begin{aligned} BD^2 &= (\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c}).(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})\\ &= \lambda_1^2 c^2 + \lambda_2^2 a^2 + 2\lambda_1 \lambda_2 \boldsymbol{a}.\boldsymbol{c} \\ &= \frac{1}{(a+c)^2} \left[a^2 c^2 + c^2 a^2 + ac(a^2 + c^2 - b^2) \right]\\ &= \frac{ac}{(a+c)^2} \left[2ac + a^2 + c^2 - b^2 \right] \\ &= \frac{ac(a+b+c)(a-b+c)}{(a+c)^2}. \quad \quad (3)\end{aligned}

Combining (2) and (3) in (1) leads to the following feast of cancellation:

\begin{aligned} \frac{DE}{EF} &= \frac{(\lambda_1 - \lambda_2) \boldsymbol{c}.(\lambda_1 \boldsymbol{a} + \lambda_2 \boldsymbol{c})}{BD^2} \\&= \frac{(a-c)}{(a+c)} \frac{a(a+b+c)(a-b+c)}{2(a+c)} \frac{(a+c)^2}{ac(a+b+c)(a-b+c)} \\ &= \frac{a-c}{2c},\end{aligned}

as was found before. Note that expressions (2) and (3) are respectively the square of the length of the angle bisector BD^2 and the dot product \vec{BC}.\vec{BD} in terms of the side lengths of any triangle, and we have found them to be similar in form.

Blog at

%d bloggers like this: