Inspired by(copied from) IGSCE

Geometry Level 4

Given that A Z = 3 cm , Z C = 2 cm , M C = 5 cm , B M = 3 cm AZ = 3\text{ cm}, ZC = 2\text{ cm}, MC=5\text{ cm}, BM=3\text{ cm} and X Y Z ∣ ∣ B M C XYZ||BMC . If the ratio of the areas of the triangles, [ A X Y ] [ A Y Z ] \dfrac{[AXY]}{[AYZ]} is a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 8.

Aareyan Manzoor by Aareyan Manzoor , 18, USA

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3 solutions

Thale's theorem in Δ A M C \Delta AMC : Y Z M C = A Z A C = 3 5 ⇒ Y Z = 5 × 3 5 = 3 c m \dfrac{YZ}{MC} = \dfrac{AZ}{AC} = \dfrac{3}{5} \Rightarrow YZ = 5 \times \dfrac{3}{5} = \boxed{3 cm} .

Thale's theorem in Δ A B M \Delta ABM : X Y B M = A X A B = 3 5 ⇒ X Y = 3 × 3 5 = 9 5 c m \dfrac{XY}{BM} = \dfrac{AX}{AB} = \dfrac{3}{5} \Rightarrow XY = 3 \times \dfrac{3}{5} = \boxed{\dfrac{9}{5} cm} .

Now, a r ( A X Y ) a r ( A Y Z ) = X Y Y Z \dfrac{ar(AXY)}{ar(AYZ)} = \dfrac{XY}{YZ} [Since they have common altitude] = 3 5 = \boxed{\dfrac{3}{5}} .

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or you could use a r e a = a b 2 sin ⁡ ( θ ) area=\dfrac{ab}{2} \sin(\theta) Nice sol(+1).

Aareyan Manzoor - 5 years, 2 months ago

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I wanted the solution to be accessible to a wider lot, so didn't involve trig here. Thanks !

Venkata Karthik Bandaru - 5 years, 2 months ago

LOL, values of A Z AZ and Z C ZC are not required!

Nihar Mahajan - 5 years, 2 months ago
Nihar Mahajan
Apr 12, 2016

Since X Z ∣ ∣ B C XZ \ || \ BC , we have X Y B M = Y Z M C = A Y A M ⟹ X Y Y Z = B M M C = 3 5 ⟹ [ A X Y ] [ A Y Z ] = 3 5 \dfrac{XY}{BM}=\dfrac{YZ}{MC}=\dfrac{AY}{AM} \implies \dfrac{XY}{YZ}=\dfrac{BM}{MC}=\dfrac{3}{5} \implies \boxed{\dfrac{[AXY]}{[AYZ]}=\dfrac{3}{5}} .

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Moderator note:

Why is the condition " X Y B M = Y Z M C = A Y A M \dfrac{XY}{BM}=\dfrac{YZ}{MC}=\dfrac{AY}{AM} " necessary? Can't we immediately do X Y ∥ B C ⇒ 3 5 = B M M C = X Y Y Z = [ A X Y ] [ A Y Z ] XY \parallel BC \Rightarrow \dfrac{3}{5}=\dfrac{BM}{MC}=\dfrac{XY}{YZ}=\dfrac{[AXY]}{[AYZ]} ?

X Y Z ∣ ∣ B M C ∴ Δ A X Y Δ A B M , Δ A Y Z Δ A M C . ∴ X Y Y Z = B M M C = 3 5 = ratio of bases of Δ s A X Y a n d A X Z . Since the heights of these triangles is same, their areas are in proportion of there bases. T h a t i s 3 5 . ⟹ a + b = 8. XYZ ~|| ~BMC~~~~~~~~~\therefore~ \Delta ~AXY\text{~} \Delta ~ABM ,~~~~~\Delta ~AYZ\text{~} \Delta ~AMC .\\ \therefore~\dfrac{XY}{YZ}=\dfrac{BM}{MC}=\dfrac 3 5=\text{ratio of bases of } \Delta s ~AXY~and ~AXZ.\\ \text{Since the heights of these triangles is same, their areas are in proportion of there bases. }\\ That ~~ is~~ \frac 3 5.~~~\implies~a+b=8.

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How did you assume XYZ ll BMC?

Silver Vice - 5 years, 2 months ago

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You are right. From the sketch we took ||. But it should have been specified.

Niranjan Khanderia - 5 years, 1 month ago

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