Problem 29. Triangle ABC has an area of 60. Point I is the midpoint of side BC and the points J and K divide AC into 3 equal segments. L is the intersection of AI and BJ. Find the area of the triangle ALJ. So let's first look for things we can compare ALJ to. Let's draw the lines LK and LC. And now we have 3 triangles, ALJ, JLK, and KLC. They all have the same height, which is the height from L to AC. And they all have equal length bases because we know that J and K trisect AC. This means they have equal areas and let's denote this area by X. Now, if we look at LIB and LCI, that's another pair of triangles with equal areas because they have the same height from L to BC and because BI equals IC. Let's call this area Y. Now we'll look at other triangles whose areas we know and try to write them in terms of sections of size X and sections of size Y in order to help us find X and Y. Remember, our goal is to find the area of triangle ALJ, which is X. So let's consider AIC. So first we see that AIC has the same height as ABC, but its base, IC, is half of the base of BC. That means its area is half of the area of ABC, so it's 30. But we can also break up AIC into LIC and AJL, JLK, and KLC. And so we can also write the area of AIC as 3X plus Y. If we look at BJC, it has the same height as ABC, the height from B to AC, and the base, JC, is two-thirds of the base, AC. This means that its area is two-thirds of the area of ABC, so BJC has area 40. We can also break it up into BLI, LIC, JLK, and KLC. So Y, Y, X, X. So we can know that BJC has area 2X plus 2Y, which is equal to 40. Now we can solve these two equations to get that X is equal to 5. Again, X is what we're looking for, X is the area of ALJ. And so our answer is B, 5.

triangle area

geometric properties

midpoint

equal parts

area calculation