COOL GEOMETRICAL FIGURE - CIRCLE

Level 2

Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumferance of the circle then 2r equals

2 P Q . R S P Q + R S \frac{2PQ.RS}{PQ+RS} P Q 2 R S \frac{PQ^{2}}{RS} P Q × R S \sqrt{PQ \times RS} P Q + R S 2 \frac{PQ+RS}{2}

Anirudha Nayak by Anirudha Nayak , 24, India

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1 solution

Anish Puthuraya
Jan 22, 2014

Keeping a general diagram in mind (or better on paper),
Let us think.

First of all, if X X lies on the circle, then it is clear that P X PX must be Perpendicular to R X RX

Hence, Let ∠ X P R = θ \angle XPR = \theta Then,
⇒ ∠ X R P = 90 − θ \Rightarrow \angle XRP = 90-\theta (By the argument mentioned before)

Also, since Both P Q PQ and R S RS are tangents to the circle,
⇒ \Rightarrow P Q PQ and R S RS are perpendicular to P R PR

Hence,
∠ Q P X = 90 − θ \angle QPX = 90-\theta
∠ S R X = θ \angle SRX = \theta

Now,writing the magnitude of P X PX in two different ways,
we have,
P X = P Q cos ⁡ ( 90 − θ ) = P R cos ⁡ θ PX = PQ\cos(90-\theta) = PR \cos\theta
⇒ P Q sin ⁡ θ = 2 r cos ⁡ θ \Rightarrow PQ\sin\theta = 2r\cos\theta ---------------------------------- Equation 1

Similarly, R S cos ⁡ θ = 2 r sin ⁡ θ RS\cos\theta = 2r\sin\theta ------------------------- Equation 2

Multiplying Equation 1 and Equation 2 ,
we get,
4 r 2 = P Q ⋅ R S 4r^2 = PQ\cdot RS

⇒ 2 r = P Q ⋅ R S \Rightarrow 2r = \boxed{\sqrt{PQ\cdot RS}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Here's a solution inspired by the fact that P R 2 = P Q ∗ R S PR^2=PQ*RS looks like a well-known formula in a particular geometric figure........:

Through R R draw a line parallel to P S PS which intersects P Q PQ at S ′ S' . Since P Q ∥ R S PQ\parallel RS (both perpendicular to P R PR ), thus P S R S ′ PSRS' is a parallelogram ⇒ R S = P S ′ \Rightarrow RS=PS' . Since ∠ P X R = 90 \angle PXR=90 , therefore ∠ Q R S ′ = 90 \angle QRS'=90 , hence P R PR is the altitude on the hypotenuse of right triangle Q R S ′ QRS' , by its well-known formula we have P R 2 = P Q ∗ P S ′ = P Q ∗ R S PR^2=PQ*PS'=PQ*RS . Q.E.D

Note: My original solution is to notice that P R P Q = R X P X , P R R S = P X R X \frac {PR}{PQ}=\frac {RX}{PX},\frac {PR}{RS}=\frac {PX}{RX} , multiply these to get our desired result.

Xuming Liang - 7 years, 4 months ago

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Cool. I thought of this problem as more of trigonometry rather than geometry. Excellent geometrical solution.!

Anish Puthuraya - 7 years, 4 months ago

Another geometric solution P X R ∼ Q P R P X = Q P . P R / Q R l l y X R = R S . P R / P S PXR\sim QPR\\PX=QP.PR/QR\\lly XR=RS.PR/PS

Also P Q 2 = Q X . Q R PQ^2=QX.QR and R S 2 = P S . X S \ RS^2=PS.XS

Also by pythagoras Q R ( Q R − R X ) = P S ( P S − X S ) = P R 2 QR(QR-RX)=PS(PS-XS)=PR^2

multiply these two and use above equations

Megh Parikh - 7 years, 4 months ago

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