Difference between revisions of "Power of a Point Theorem/Introductory Problem 3"

Revision as of 08:27, 1 July 2006

Problem

(ARML) In a circle, chords $AB$ and $CD$ intersect at $R$ . If $AR:BR = 1:4$ and $CR:DR = 4:9$ , find the ratio $AB:CD.$

Solution

Letting $AR = x$ makes $BR = 4x$ . Similarly, letting $CR = 4y$ makes $DR = 9y$ . Thus $AB = AR + BR = x + 4x = 5x$ and $CD = CR + DR = 4y + 9y = 13y$ . We therefore seek $\frac{AB}{CD} = \frac{5x}{13y}$ .

From the Power of a Point Theorem we have that

$x\cdot 4x = 4y\cdot 9y\Rightarrow \left(\frac xy\right)^2 = 9$

which gives $\frac xy = \pm 3$ so we take $\frac xy = 3$ .

Finally

$\frac{5x}{13y}=\frac 5{13}\cdot \frac xy = \frac 5{13}\cdot 3 = \frac{15}{13}.$

Back to the Power of a Point Theorem.

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Difference between revisions of "Power of a Point Theorem/Introductory Problem 3"

Revision as of 08:27, 1 July 2006

Problem

Solution

Revision as of 06:40, 1 July 2006 (view source) Joml88 (talk \| contribs) m (Power of a Point Introductory Problem 3 moved to Power of a Point Theorem Introductory Problem 3) ← Older edit	Revision as of 08:27, 1 July 2006 (view source) Joml88 (talk \| contribs) m (Power of a Point Theorem Introductory Problem 3 moved to Power of a Point Theorem/Introductory Problem 3) Newer edit →
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