## ________________

*MoMath Masters*tournament, held in Tribeca (the TRIangular area BElow CAnal street) in Manhattan. The students congratulate the coach by posing some problems about areas related to triangles.

## ________________

### Trial Squares

Jess experiments with two different ways that a square can be inscribed into the same isosceles right triangle: In one way, two of the square’s sides lie exactly along the two legs of the triangle, and one of its vertices just touches the hypotenuse. The other way, one of its sides lies exactly along the hypotenuse, and two of its vertices just touch the legs of the triangle. For the triangle under consideration, Jess notices that the area of the larger of these two squares is 576.

*What is the area of the smaller of the two squares?*

### Four of Six

Triangle *ABC* is divided into six smaller triangles by three line segments from the vertices, all through the same interior point, as shown. Some of the areas of the smaller triangles are given in the diagram.

*What is the area of triangle ABC?*

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## Solutions to week 76

**Unleash Your Inner Rectangle**. The accompanying figure shows quadrilateral *ABCD* with midpoint rectangle *EFGH*, as well as point *K*, the intersection of the two diagonals *AC* and *BD*. Since *AH* is half of *AD* and *AE* is half of *AB*, triangle *AHE* is similar to triangle *ADB*. That means that angle *AHE* is equal to angle *ADB*, so lines *HE* and *BD* are parallel. Reasoning similarly about triangles *BEF* and *BAC*, we conclude that lines *EF* and *AC* are parallel. Since *HE* and *EF* are adjacent sides of a rectangle, they are perpendicular, so *BD* and *AC* are perpendicular as well. In particular, triangle *ABK* is a right triangle; since one acute angle *CAB* is given as 17 degrees, the other one *ABD* is **73 degrees**.

**Pent-up-gon**. The diagram shows pentagon *ABCDE* with midpoint regular pentagon *FGHJK*, as well as auxiliary lines *AC, CE, EB, BD*, and *DA*. Reasoning about similar triangles just as in the previous problem, we see that these auxiliary lines are each parallel to, and twice as long as, *FG, HJ, KF, GH*, and *JK*, respectively. Thus all of the auxiliary lines are equal, and given they make the same angles to each other as the sides of the regular pentagon *FGHJK* do, the entire dashed figure is a regular pentagram. But the vertices of a regular pentagram form a regular pentagon, so *BC* = *CD*, and the ratio asked for is **one**.

## Recent Weeks

**Week 76**: Unleash Your Inner Rectangle & Pent-up-gon, solutions to Some Difference & In Your Prime

**Week 75**: Some Difference & In Your Prime, solutions to Frequent Figures & Doubled Digit

**Week 74**: Frequent Figures & Doubled Digit, solutions to Small Sudoku & Puny Pegs

**Week 73**: Small Sudoku & Puny Pegs, solutions to The Second Bisection & The Third Circle

**Week 72**: The Second Bisection & The Third Circle, solutions to Eye Like It & Pythagorean Poster

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

**Come back next week** for answers and more puzzles.