Geometry Hustle 2019 MA National Convention
Geometry Hustle Answers
1. 525𝜋
2. 441𝜋
3. 8
4. 4
5. 125/2 or 62.5 or 62
6. 17/2 or 8.5 or 8
7. 35
8. 5
9. The open interval (3, √65)
10. 65/2 or 32 or 32.5
11. 1/3
12. 30
13. √3 + 2√2 or 2√2 + √3
14. 8
15. 4
16. 7
17. 1/4 or 0.25
18. 7
19. 5
20. 27
21. 3
22. abcd
23. Proof by Construction or Constructive Proof
24. 4
25. 5
Geometry Hustle 2019 MA National Convention
Geometry Hustle Solutions
(1) 𝑽 = 𝝅𝒓𝟐 𝒉. Plugging in the values gives a volume of 𝟓𝟐𝟓𝝅.
(2) As 𝑛 → ∞, the 𝑛-gon approaches a circle, thus the apothem length is the radius of the
circle, so the area approaches 441𝜋.
(3) It is clear that the maximum with 3 planes is 8 regions.
(4) 𝐻𝐵 = 4 by power of a point.
(5) Applying the formula 𝑎𝑏 sin 𝐶, we get that [𝐾𝐼𝑀] = (10)(25) sin 150∘ =
(10)(25) = .
(6) By Shoelace Formula, the area is 17/2.
(7) Aaron must take 4 steps up and 3 steps right, which gives = 35 total moves.
(8) By triangle inequality on 𝑃𝐴𝐶, we see that 6 < 𝑃𝐶 < 14. By triangle inequality on
𝑃𝐶𝐾, we see that 1 < 𝑃𝐶 < 13. Thus, 6 < 𝑃𝐶 < 13, which gives max(𝑃𝐶) = 12 and
min(𝑃𝐶) = 7, 12 − 7 = 5.
(9) By triangle inequality, 𝑥 > 3. By Pythagora’s Inequality, 𝑥 < √65. Thus (3, √65).
(10) The area of the triangle is (5)(13) sin 𝐾 where angle 𝐾 is across from side length 𝑘.
The constraint is simply the triangle inequality. Thus the maximum area is when
sin 𝐾 = 1 or when this is a right triangle. The area is 65/2.
(11) Without loss of generality, let 𝐶 lie on the positive 𝑥-axis, 𝐴 lie on the positive 𝑦-axis,
and 𝑀 lie on the positive 𝑧-axis. The satisfactory tetrahedron has vertices 𝐶 = (1,0,0),
𝐴 = (0,2,0), and 𝑀 = (0,0,1). The volume is base [𝐶𝑂𝑀] = 0.5 times height 𝑂𝐴 = 2
over 3, which evaluates to 1/3.
(12) An equilateral quadrilateral is a rhombus. A rhombus with diagonals 5 and 12 has area
30.
(13) [𝑆𝑈𝑁] is 2√3 with probability 1/2 and 4√2 with probability 1/2. Thus the expected
value of [𝑆𝑈𝑁] is √3 + 2√2.
(14) Note that 𝑁𝐸 can take on all lengths from 0 to 8. To visualize this, take the tangent from
𝑆 to circle 𝐼. This is when 𝑁 and 𝐸 are the same point (𝑁𝐸 = 0). Moving 𝐸 along the
circle to 𝐻′ such that 𝐻𝐼𝐻′ is a diameter (𝑁𝐸 = 8), we see that the length of 𝑁𝐸 grows
in a continuous manner between these endpoints. Note that 𝐼𝑁 = 𝐼𝐸 = 4, so 𝐼𝑁𝐸 is
isosceles. Thus we maximize the value of (𝐼𝑁)(𝐼𝐸) sin ∠𝑁𝐼𝐸 = 8 sin ∠𝑁𝐼𝐸. The
maximum value for sine is 1, which occurs when 𝑚∠𝑁𝐼𝐸 = 90∘ or 𝑁𝐸 = 4√2, which
is valid since 0 < 4√2 < 8.
(15) Let the leg lengths of a right triangle be 𝑎, 𝑏 and hypotenuse be 𝑐. The romi ratio is
( )
= . Note that the AM-GM inequality states that for non-negative numbers
/
Geometry Hustle 2019 MA National Convention
𝑥, 𝑦: ≥ 𝑥𝑦. This is equivalent, for 𝑥, 𝑦 ≠ 0, ≥ 2. Let 𝑥 = 𝑎 and 𝑦 = 𝑏 , and
√
( )
apply AM-GM to get ≥ 2 which multiplying both sides by 2, we get ≥4
as desired.
(16) Plugging in the values into Apollonius’s Identity gives the following. 7 + 9 = 2𝑥 +
. Solving for 𝑥 gives 𝐾𝑁 = 7.
(17) [𝑆𝑅𝑇] is half the area of 𝑆𝑇𝑌 since they share an altitude. [𝑅𝑂𝑇] is half the area of 𝑆𝑅𝑇
[ ]
by the same logic. Thus [ ] = .
(18) 𝐾𝑁 is the hypotenuse of right triangles 𝐾𝑂𝑁 and 𝐾𝐺𝑁. Let 𝐾𝐺 = 𝑥. 𝑥 + 36 = 4 +
81 ⟹ 𝑥 = 7.
(19) Let 𝐸𝑉 lie tangent to the semicircle at point 𝐴. Since tangents from the same point have
equal length, 𝐸𝐴 = 𝐸𝐹 = 4. Additionally we can express 𝑆𝑉 = 𝑉𝐴 = 𝑛. Thus 𝑉𝑇 =
4 − 𝑛. Using Pythagorean Theorem on 𝑉𝐸𝑇, we get 𝑛 − 8𝑛 + 16 + 16 = 𝑛 + 8𝑛 +
16. Solving, we get 𝑛 = 1. We are solving for 𝐸𝑉 = 𝑛 + 4 = 5.
(20) By similar triangles, 𝐻𝑀 = 9. Since 𝐻𝐴𝑀 is a right triangle, [𝐻𝐴𝑀] = (6)(9) = 27.
(21) i. (Squaring a circle) and iii. (Trisecting an angle), are 2 canonical examples of
impossible constructions, everything else is possible.
(22) This is a biconditional, so by definition 𝑋 → 𝑌 and its converse must be true. Since the
inverse is the contrapositive of the converse, and contrapositives hold the same truth
value as the statement, the inverse of 𝑋 → 𝑌 must be true. Thus, all abcd must be true.
(23) This is the definition of a proof by construction or constructive proof.
(24) All squares and rectangles are cyclic. An equilangular rhombus is the same thing as a
square. Isosceles trapezoids are cyclic. Only the kite is not necessarily cyclic. Therefore,
i, ii, iii, iv are cyclic.
(25) The tangent implies that 𝑆𝑅𝑌 is a right triangle, so 𝑆𝑇𝑂𝑅 is inscribed in a semi-circle
with radius 5. Thus, 𝑆𝑇𝑂𝑅 is an isosceles trapezoid, so 𝑇𝑋 is equal to the radius, which
is 5.
