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Well done to Avan from Tanglin Trust School and Kyle in Singapore, Theo from Pate's Grammar School in the UK and Mahdi from Mahatma Gandhi International School in India who all sent in correct proofs. This is Avan's work to prove that triangle $BFC$ is congrugent to triangle $DAC:$
Similarly, Theo wrote:
All interior angles on a regular pentagon are 108˚. Therefore, angles CBF and BCF are 72˚, as when put with angles ABC/BCD, they make a straight line. That means angle BFC is 36˚ and triangle BCF is isosceles. Next, we know that triangle ABC is isosceles as lines AB and BC share the same length. Knowing angle ABC is 108˚, angles ACB and CAB are both 36˚. With this, angle ACD is 72˚, and so is ADC. The bases for triangles BCF and ACD are the same length, and the two equal angles are the same in both triangles, so the two triangles are the same (angle side angle theorem).kyl
Mahdi, Kyle and Avan all found a pair of similar triangles. Kyle and Avan found the same pair. This is Kyle's work:
Kyle and Avan found the length DA. Avan used trigonometry (click below to see the text in Avan's work):
With our new triangle EDX we have the length of the hypotenuse (1) and the angle XED (54). This is because to get angle XED you had to divide angle DEA by 2, which is 108/2, which is 54.
On this triangle the length XD is the opposite side, while the length ED is the hypotenuse. We can solve for the length of XD using the equation sin(54) = opp/hyp.
This can be rearranged to give us hyp$\times$sin(54)=opp
We can substitute in the hypotenuse length (1) to get sin(54)=opp
sin(54) = 0.809 (3 s.f.)
We can multiply this by 2 to get the full length of AD, 0.809$\times$2=1.618
Kyle and Mahdi used their similar triangles to find the length DA. This is Kyle's work:
Mahdi found and used a different pair of similar triangles to find DA. Mahdi's work is similar to Kyles, but is particularly neat:
