GERMANY MATH OLYMPIAD GEOMETRY CHALLENGE

[00:00:00]What's up everyone? Welcome back.

[00:00:03]Today's challenge comes from the German math Olympiad. It's a beautiful geometry problem that has puzzled a lot of students. The trick isn't complicated calculations, it's spotting the right construction. Let's see if we can solve it together.

[00:00:20]Let's begin with the figure. We have triangle ABC.

[00:00:25]Point P lies on the base BC, dividing the base into two segments.

[00:00:30]Now let's look at the information we're given. Side AB has the same length as segment CP.

[00:00:38]For convenience, let's call that common length A. So AB equals A and CP also equals A.

[00:00:47]We're also given three angles. Angle PAC is X, angle PCA is 2X, and angle PBA is 10X.

[00:00:59]Our goal is to find the value of X.

[00:01:02]At first glance, there doesn't seem to be enough information. So instead of trying to force a solution, we're going to create one.

[00:01:12]Here's the clever construction.

[00:01:15]From point P, draw a new segment to a point Q so that angle APQ is exactly X.

[00:01:25]This one extra line unlocks the entire puzzle.

[00:01:30]Now look at triangle APQ.

[00:01:33]Angle APQ is X.

[00:01:36]And angle PAQ is also X.

[00:01:40]Since two angles are equal, this triangle is isosceles.

[00:01:45]Therefore, AQ equals PQ.

[00:01:49]Now use the exterior angle theorem. The exterior angle at Q equals the sum of the two remote interior angles.

[00:02:00]So, the exterior angle is 2x.

[00:02:05]Now, turn your attention to triangle CPQ.

[00:02:09]Angle PCQ is already 2x, and we just found that angle PQC is also 2x.

[00:02:17]So, this triangle is isosceles as well.

[00:02:21]That means CP equals PQ.

[00:02:26]But, CP was given to be A.

[00:02:29]So, PQ equals A.

[00:02:33]And since AQ equals PQ, we also have AQ equals A.

[00:02:41]Now, we've shown that both AB and AQ have length A.

[00:02:47]Next, draw the segment BQ.

[00:02:52]Now, let's look at the large triangle ABC.

[00:02:57]The angle at B is 10x.

[00:03:01]The angle at C is 2x.

[00:03:04]So, the angle at A must be 180° 12x.

[00:03:13]Now, focus on triangle ABQ.

[00:03:18]We already know that AB equals AQ.

[00:03:22]So, this triangle is also isosceles.

[00:03:26]Let's call each base angle theta.

[00:03:30]>> [snorts] >> Adding the three angles gives 180° - 12x + theta + theta = 180°.

[00:03:45]Simplifying, 2 theta = 12x.

[00:03:50]So, theta equals 6x.

[00:03:55]That means the angle at B inside triangle ABQ is 6x.

[00:04:04]But the entire angle PBA is 10x.

[00:04:10]So, the remaining angle between PB and QB must be 4x.

[00:04:18]Now, look at triangle CPQ.

[00:04:22]Its two interior angles are both 2x.

[00:04:26]So, the exterior angle at P is 4x.

[00:04:32]Finally, consider triangle BPQ.

[00:04:37]Angle PBQ is 4x.

[00:04:41]Angle BPQ is also 4x.

[00:04:44]So, triangle BPQ is another isosceles triangle.

[00:04:49]Therefore, BQ equals PQ.

[00:04:55]But we already proved that PQ equals A.

[00:04:59]So, BQ also equals A.

[00:05:04]Now, look back at triangle ABQ.

[00:05:07]AB equals A, AQ equals A, and BQ equals A.

[00:05:14]All three sides are equal.

[00:05:16]So, triangle ABQ is equilateral.

[00:05:21]Every angle in an equilateral triangle measures 60°.

[00:05:26]Earlier, we proved that each base angle of triangle ABQ is 6x.

[00:05:32]So, 6x equals 60°.

[00:05:38]Divide both sides by six, and we get x equals 10°.

[00:05:45]One carefully chosen construction creates a chain of isosceles triangles.

[00:05:51]And that chain leads us all the way to an equilateral triangle.

[00:05:56]That's the hidden idea behind this beautiful Olympiad problem.

[00:06:02]>> [snorts] >> Did you spot the construction before the proof began? Let me know in the comments.

[00:06:06]Thanks for watching, and I'll see you in the next video.