Problem 27 · 2014 Math Kangaroo
Stretch
Geometry & Measurement
symmetry
PT is tangent to a circle with centre O, and PB is the bisector of the angle TPA (see diagram). How big is the angle TBP?
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Answer: B — 45°
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Hint 1 of 2
Use the tangent–chord angle and the fact that PB bisects angle TPA.
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Hint 2 of 2
Chase the angles to show angle TBP does not depend on where P sits.
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Approach: tangent–chord angle plus the bisector make the P-dependence cancel
- Let \(\angle TPA = 2\alpha\), so the bisector gives \(\angle TPB = \alpha\). By the tangent–chord angle, the angle between tangent \(PT\) and chord \(TB\) equals the inscribed angle \(TB\) subtends, namely \(\angle TPB + \angle PBT = \alpha + \angle TBP\) seen from the alternate segment.
- Writing the angle sum of triangle \(PTB\) and substituting the tangent–chord relation, every term involving \(\alpha\) (hence the position of \(P\)) cancels.
- What remains forces \(\angle TBP = \) 45°, the same for every position of \(P\).
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