Many years ago, having heard of its mighty powers in healing a terrifying disease, a cutie pie schoolgirl sought out a Temple in Novena, together with her schoolmates.

Upon discovering that the High Priestess of this temple also possessed the ability to foretell the future, they burst into an uncontrollable delirium of giggles, as teenage girls tend to do, regardless of whether they’re really cute or not.

And so after being cajoled by her bunch of bitchy-looking Eurasian schoolmates, the schoolgirl entered the temple to stand alone before its ultra-sexy High Priestess.

“*How will my future partner be?*” she asked the High Priestess.

Lei gor mei loy lam yan wui ho leng chai.

Lei yun fan wui sam yuet sup sei hou yat dim mmm sup gao fen dou.

replied the High Priestess in Cantonese, which was loosely translated as

“*Your future man very the handsome, and you will find your fate on March 14 at exactly 1.59pm.*”

With that, the High Priestess shoved a folded piece of paper into her hand, telling her not to open it till she had found The One on a March 14th , else she would remain a spinster all her life and die horrible deaths in all her future maths tests forever ever after.

Fast forward to the present (March 14 to be exact), this cutie pie schoolgirl was waiting in one of those never ending queues in one of Singapore’s world-class bus interchanges when she caught a glimpse of a gorgeous hunk standing just a couple of places ahead of her.

Like the sunshine which had lifted the gloomy weather in the past days, her face brightened as she slowly surveyed his long flowing hair that resembled infinite tangent curves, and his well-chiseled face that resembled some n-sided polygon.

Her heart skipped a beat when she spotted the cool **π** tattoo on his shoulder, and that he was holding a file containing the same set of exam papers as hers.

The moment of realization was complete when her watch showed 1.59pm, and 26 seconds.

Unfolding frantically the tattered piece of paper that she had kept all these years, she saw the following …

An old Pi Day tradition allows a woman to make the first move in a romance, provided that she manages to touch a man’s

π-spot at exactly 1.59pm, and the man shall turnirrationalenough to accept her.The vicinity of a man’s

π-spot can be found by solving the problem below, withObeing the location of his heart.

In the figure, the sector `OPQ` has centre `O` and radius 6 cm and ∠`POQ` = 36^{o}. `OR` bisects ∠`POQ` and `S` is the midpoint of `OR`. An arc, centre `S` and radius 3 cm is drawn to meet `OP` and `OQ` at `U` and `V` respectively.

- Find, in terms of
`π`,- the length of arc
`PRQ`. - the area of sector
`OPQ`. - the size of ∠
`OSU`.

- the length of arc
- Find the area of the shaded region
`PRQVU`correct to 3 significant figures.

As our love-struck cutie π schoolgirl has not taken A-Maths before, and that Circular Measure has just invaded the Mensuration chapter of the new E-Maths Syllabus, can you help her find the hunk’s **π**-spot?

There’re not many seconds left before 2.00pm!

Multi-part *cheong hei* aka long-winded questions like this can often catch out weary students during exams. For they *ought* to know that:

The length of an arc, `s` = `rθ`,

and that the area of a sector, `A` = `r`^{2}`θ`,

and that `θ` must always be in *radians*,

and that `π` radians = 180^{o},

and not forgetting something that Miss Loi has already highlighted before here, amongst other things.

The situation is not helped when sadistic examiners sometimes ‘conveniently’ forget to include in the diagram (maybe they’re drawn by someone else?) certain information described in the question. Please save yourself from a major eyeballing exercise by writing down the information on the diagram as soon as you see it!

P.S. This is supposed to be the Pi Day post on 14 March, but Miss Loi has been busily choking on intensive *joss sticks* smoke this whole week.

P.P.S. Please no naughty comments concerning this **π**-spot! Yes **HAM**, Miss Loi’s referring to you!!!!

## 10 Comments

曜

日

Whoa really got Pi day ah..lol..n nice use of the link to the leap day tradition...

曜

日

Aiyoh

Soupwhat kind of indices expression is that? *blushes*曜

日

Liddat also can! !! So was this your love story?

曜

日

HighwayBlogger, if only it was THIS easy ... tattoo shops' business will skyrocket the week before every March 14 (just like the florists on Feb 14),~~and guys will be sitting around everywhere waiting for the ... ummm ... 'poke'~~*sorry that doesn't sound right*And SDU would've gone extinct much earlier!

曜

日

Ehhhhh!!!! What notty commrnt arrrr??? This Pie tattoo got notty spot meh??? Where arrrr???

曜

日

Easy there

HAM... easy ... it's just an innocentπ!曜

日

Hm..left unsolved for half yr...

* Our cutie π schoolgirl bemoans half a year of wasted youth :'( *

It's always a good first thing to include all values in the question in the diagram, as you may be able to see things that sometimes others can't see *hair stands on neck*

1.i.

Angle POQ = 36π/180 = 0.2π rad

(Yup - just wish to reiterate once more that 180° =

πradians NOT 2π!)Arc PRQ = PO x 0.2π = 6 x 0.2π = 1.2π cm

(Yup - simple substitution into arc length formula

s=rθ, with radiusr= 6 cm andθ= 0.2π)ii.Area of sector OPQ = ½(6^2)(0.2π) = 3.6π cm

^{2}(simple substitution into sector area formula formula , with radius

r= 6 cm andθ= 0.2π)No sweat so far right? Remember again that all the

θin your arc and sector area formulae must be inradiansi.e. in terms ofπand/or switch your calculators into RAD mode.iii.Angle SOU = Angle SOV = 0.2π/2 = 0.1π rad

Angle OSU = π (ie 180 °) - (2x0.1π) = π - 0.2π

= 0.8π rad

Part 2 coming up later...

Yaloras seen from the diagram and described in part 2 of your workings below, since:OS= 3cm (asSis the mid-point ofORwhich is 6cmUS= 3cm (since it's givenSis the center of arcURVwith radius 3cm)So

OS=US⇒ Δ

USOis isosceles⇒ ∠

SUO= ∠SOU= 18°⇒ ∠

OSU= 180° - 18° - 18° = 144° = 0.8πradSee your next comment below for an alternate approach ...

曜

日

Full correction to 1iii.

SU = SR (radii, centre S)

= OS (S midpt of radius OR) = 6/2 = 3cm

Since SU = OS, triangle OSU is isosceles and thus

angle SUO = angle SOU = angle POQ/2

= 0.2π/2 = 0.1π rad

Angle OSU = π(sum of angles in triangle) - (0.1πx2)

= π - 0.2π = 0.8π rad

So tedious...

Yalorvery tedioushor?If one looks hard enough at the diagram, since

US=OS=VS, a circle with centerSwill magically materialize:And using one of the

Angle Properties of Circles(which surfaced in the little link Miss Loi included in the main article) where:∠ at center = 2 × ∠ at circumference subtended by the same arcSo we get ∠

USV= 2 × ∠UOV= 0.4πrad∴ ∠

OSU=π- (∠USV)/2 (since it's givenORbisects ∠POQ) = 0.8πrad 😉曜

日

Yes to find the shaded area, your approach is summarized as follow:

Area of Sector

OPQ(= 3.6πcm^{2}obtained in 1ii. above)- Area of Sector

USV- Area of Δ

OSU- Area of Δ

OSV2. By symmetry, angle OSV = angle OSU = 0.8π rad

Angle USV = 2π(angles at a point) - (2 x 0.8π)

= 2π - 1.6π = 0.4π rad (which is obtained directly if one has used the 2nd method of calculating ∠

OSUabove - but follow whichever method you're comfortable with)Area of sector USV = ½ (SU

^{2})(0.4π)= ½X9X0.4π =1.8π cm

^{2}Again by symmetry, area of triangle OSU

= area of triangle OSV

= ½(OSxSU)xsin0.8π

(using the ½

ABsinθarea of Δ formula you've learned in your Sine Rule chapter)= ½x9xsin0.8π = 2.645cm

^{2}(4 sig fig, from calculator)

So shaded area PRQVU = sectorOPQ - sectorUSV

- trianglesOSU&OSV

= 3.6π - 1.8π - (2.645x2)

= 1.8π - 5.29 = 0.365cm

^{2}(YEAH!)(using calculator π, rounding off ans to 3 sig fig)

It's long past 2pm... The cutie π girl is very touched by you burning the midnight oil for her! <3<3<3

曜

日

Thanks to

WS, may our cutie π girl finally find her true love on March 14 1.59pm2009.*Wonders where would

WSbe on that day* 😛