A Very Nice Math Olympiad Geometry Challenge 

BD and CE are medians, so F is centroid. Then height of BCF = 1/3 times height ABC, [BCF] = 480 / 3 = 160. BP/BD = 1/2, BF/BD = 2/3 => BP / BF = 3/4 => PF = BF / 4. Same reasoning for FQ; FQ = CF / 4. Then [PFQ] = [BCF] / 4^2 = 160 / 16 = 10

Good solution, however, I made it slightly different as follows: I called Area [FPQ]=S From the information that was given, we can notice that CE and DE are medians and point F is the Centroid. So, the segments BF and FD have a relation of 2:1. Let´s say that DF=2a, then BF=4a. However, BP=DP and because of that, we can say that PF=a e BP=3a. Using that same analogy for median CE, we can say that EF=2b, FQ=b and CQ=3b. I drew a segment EP that divides triangle EPQ into two triangles EFP and PFQ, which have area relation of 2:1 based on the segments EF=2b and FQ=b. Since I called Area [FPQ]=S, then Area [EPF]=2S Now let´s look at triangle EBF that is divided buy the cevian EP, which divides area [EBP] and area [EPF] with the relation 3:1 based on segments BP=3a and PF=a. Since Area [EPF]=2S, then area [EBP]=6S. Total area[EBF]=8S Now let´s look at triangle EBC that is divided buy the cevian BF, which divides area [EBF] and area [FBC] with the relation 2:1 based on segments EF=2b and FC=4b. Since Area [EBF]=8S, then area [FBC]=16S. Total area [EBC]=24S Now let´s look at triangle ABC that is divided buy the median CE, which makes area [EBC] the same as area [ECA] based on segments BE=EA. Since Area [EBC]=24S, then area [ECA]=24S Area [ABC]=480=48S. So area[FPQ]=S=10

📝 Summary of Key Points: The video explains the relationship between different points and areas in a triangle, including the medians AE and BE intersecting at point G, and the ratios AG:GD, BG:GE, and CG:GF being 2:1. The video calculates that the area of QFP is 1/4 of the area of EFD based on the similarity of triangles EDP and QFP. The ratio of the areas of two triangles is equal to the ratio of their corresponding sides squared, as demonstrated by the area of ABD divided by the area of ACD being equal to BD:CD. The area of QFP is determined to be 10, leading to the conclusion that the area of FPQ is also 10. 💡 Additional Insights and Observations: 💬 "If we join DE, the angles at point F in triangle EDP and triangle QFP will be equal." 📊 The area of ABC is given as 480, and the area of ABD is calculated to be 240. 🌐 The concept of the ratio of areas of triangles being equal to the ratio of their corresponding sides squared is a fundamental property in geometry. 📣 Concluding Remarks: The video provides a clear explanation of the relationship between different points and areas in a triangle. By utilizing the properties of medians and the ratio of areas to corresponding sides, the video successfully calculates the area of FPQ as 10. This demonstrates the importance of understanding geometric concepts and their applications in problem-solving. Generated using Talkbud (Browser Extension)

Since BD and CE are medians, [BCD] = [BCA] = 480/2 = 240, let BF= 2x, FD= x, CF = 2y, EF = y, then PD = 3x /2 = 1,5x and PF = PD- FD = 1,5x- x = 0,5 x and EQ = 3y/3 = 1,5y and FQ = EQ - EF = 1,5y - y = 0,5y The ratio [FCD] / [BCD] = ((1/2)*x*CD) / ((1/2)*3x*CD = ((1/2)/(3/2) = 3 Therefore the area [FCD] = [BCD]/3 = 240 /3 = 80 Also the area [BCF] = [BCD] - [FCD] = 240 - 80 = 160 The ratio [PQF] / [BCF] = (((1/2)*PF*FQ*sinθ))/ ((1/2)*BF*CF*sinθ)) = (0,5x * 0,5y) / (2x*2y) =1/16 Therefore [PQF] = (1/16)* [BCF] = (1/16) * 160 = 10

Very Nice!

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I have a math problem. I am unable to solve it. Will you solve it for me

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