AMC 12 Geometry: In a Secant Product, Use the Whole Secant Length

A circle diagram can contain several lengths on one line. Power-of-a-point relationships use particular products, so labelling external and whole lengths prevents a common substitution error.

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Specify the configuration

From a point P outside a circle, a secant meets the circle first at A and then B. Suppose PA = 9 and PB = 16. A tangent from P touches the circle at T.

The tangent-secant theorem gives PT2 = PA × PB = 144, so the tangent length is PT = 12.

Identify the internal segment

The inside part AB is PB - PA = 7. It is not the second factor in this theorem. Using 9 × 7 would confuse the whole secant with its internal part.

All lengths must refer to the same external point P and the actual circle configuration.

Check applicability before multiplying

If the point were inside the circle, the diagram and relevant segment relationship would need reinterpreting. A theorem name alone does not establish its conditions.

Mark P-A-B in order before writing the product, and choose the positive root because a segment length is positive. This original example supports AMC 12 geometry reasoning and is not a released MAA problem.

Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.

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Official information and further reading: Source 1

AMC 12 Probability: Expected Counts Can Add Even When Events Overlap

When a random quantity counts many local events, indicators can simplify its expectation. The events may overlap, but expectation still adds. Independence is needed for some probability products, not for linearity itself.

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Define exactly what is counted

Toss a fair coin independently eight times. There are seven adjacent pairs of positions: 1-2 through 7-8. Count how many pairs show equal results, allowing overlaps.

For any specified pair, HH and TT account for two of four equally likely outcomes, so its probability of equality is 1/2.

Add the expected contributions

Let each pair contribute 1 when equal and 0 otherwise. Each indicator has expectation 1/2. The total expected count is therefore 7/2 = 3.5.

A single realised count is an integer. An expectation of 3.5 is an average over repeated outcomes, not a claim that one sequence contains half a pair.

Do not overstate the conclusion

Adjacent-pair events share tosses, and a separate distribution calculation would need careful dependence analysis. The expectation argument does not by itself give the probability of exactly three equal pairs.

This original exercise develops probability reasoning for AMC 12 preparation, not an official problem prediction. Define the counted objects and whether overlap is allowed before introducing indicators.

Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.

Contact Hanlin

Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 EXPECTATION 1024. Include your school year, country or competition region, experience and the question you want to discuss.

Scan to contact Hanlin on WeChat. Send AMC12 EXPECTATION 1024.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1

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