Math IOQM

Question: In a triangle ABC, the median from B to CA is perpendicular to the median from C to AB. If the median from A to BC is 30, determine $$ \frac{BC^2 + CA^2 + AB^2}{100}. $$  Answer: 24 $$ AR = 30 $$ $$ AG : GR = 2:1 \\ $$ $$ GR = 10 \\ $$ $$ \text{in } \triangle BGC, \angle G = 90^\circ \\ $$ $$ BR = RC = GR = 10 \\ $$ $$ AB^2 + AC^2 = 2 (AR^2 + BR^2) \\ $$ $$ AB^2 + AC^2 = 2 (900 + 100) \\ $$ $$ AB^2 + AC^2 = 2000 \\ $$ $$ \frac{AB^2 + AC^2 + BC^2}{100} = \frac{2000 + 400}{100} = 24 $$

IOQM 2023 Q5: In a triangle ABC , let E be the midpoint of AC and F be the midpoint of AB . The medians BE and CF intersect at G . Let Y and Z be the midpoints of BE and CF respectively. If the area of triangle ABC is 480 , find the area of triangle GYZ . Answer 10:  Solution: 1. 3 medians intersecting at centroid divide the triangle into 6 triangles of equal area. 2. GBC has 2 of those triangles so its area is 1/3 of the total area. 3. Look at GYZ. GB is 2/3 of the median BE. BY is 1/2 of the median BE. So GY = 2/3 - 1/2 = 1/6BE. GYZ and GBC are similar with side ratio of 1:4. So [GYZ] = 1/16[GBC] = 1/16 * 1/3 [ABC] = 480/16*3 = 10. Prerequisites: Centroid and median properties in triangle . Similarity criterion for triangles.