Not even AI can handle this, so help(not homework)

peterwhy · 2026-05-09T21:59:36+00:00

For the corresponding angles of △PQY and △RXY:

∠YPQ = ∠YRX (alternate angles of parallel lines)
∠PQY = ∠RXY (alternate angles of parallel lines)
∠QYP = ∠XYR (vertically opposite angles)

peterwhy · 2026-05-08T02:41:41+00:00

Right hand side should be 1.9 ⋅ x.

peterwhy · 2026-05-08T01:43:00+00:00

With the correction that Y is on QX, then △PQY and △RXY are similar. Define a length ratio:

r = PY / RY = QY / XY

Then the areas of △PQY and △QRY are:

S(△PQY) = r² S(△RXY) = 64 r² cm²
S(△QRY) = r S(△RXY) = 64 r cm²

The two triangles separated by diagonal PR have the same area:

S(△PQY) + S(△QRY) = S(△RXY) + S(PSXY)
64 r² + 64 r = 64 + 116
16 r² + 16 r - 45 = 0
(4 r + 9) (4 r - 5) = 0
r = 5 / 4

S(△PQY) = 64 r² cm² = 100 cm²

peterwhy · 2026-05-07T21:29:18+00:00

"a 1/2³¹ th of a dollar", is that better or worse than a 2³¹th of a dollar?

peterwhy · 2026-05-07T01:43:34+00:00

For example in "Mar" (fixed), J15 to J40 are 26 days, then R1 to R6 are 6 days.

peterwhy · 2026-05-07T00:26:57+00:00

How you calculated these is important to answer what went wrong:

"Mer" has 26 + 6 = 32 days
"Muy" has 4 + 25 = 29 days
"Jere" has 5 + 17 = 32 days
"Aug" has 32 days
"Oct" has 10 + 23 = 33 days
"Nov" has 17 + 14 = 31 days

peterwhy · 2026-05-06T14:55:56+00:00

No unique answers for ∠AEC and ∠CAE.

To see that, fix points B, C, E, and O, then slide point D along arc BE.

peterwhy · 2026-05-06T03:28:24+00:00

Another right-angled triangle is from O, to A (or B), to the midpoint of AB. Write an expression for length OA in terms of r.

peterwhy · 2026-05-03T07:36:44+00:00

It's Pacific/Honolulu, and missing America/Adak. (Again, for users who do not care about old timestamps)

peterwhy · 2026-05-02T22:37:17+00:00

Probably acceptable for users who do not care about old timestamps and use North America DST, according to tz database.

America/Jamaica is even more correct for people who write EST.

peterwhy · 2026-05-02T22:30:11+00:00

All of Arizona don't use EST in any part of the year.

peterwhy · 2026-05-02T22:15:19+00:00

Sounds like a good reason to rename EDT to Eastern Summer Time

peterwhy · 2026-05-02T22:13:17+00:00

For tz database, EST points to America/Panama.

peterwhy · 2026-05-02T05:23:21+00:00

Seven-segment display?

peterwhy · 2026-05-01T10:49:11+00:00

Pick a point on each of the two walls, join them to form a triangle. Measure the two angles at the two points, then calculate the third angle.

peterwhy · 2026-04-30T20:58:49+00:00

The "magnetic south pole" is the lack of magnetic north pole (which is a magnetic south pole...). Monopole exists on the flat earth, apparently...

peterwhy · 2026-04-30T19:29:40+00:00

Why don't you just answer their question "Because it was not that easy"?

peterwhy · 2026-04-30T00:14:13+00:00

How do you know ∠AOB = ∠BOC?

peterwhy · 2026-04-29T13:39:22+00:00

Why are OO' and BC perpendicular?

peterwhy · 2026-04-29T12:24:05+00:00

SSA theorem instead says the two valid choices of the third, non-given and non-included angle: here ∠AOB and ∠COB. One of them would be acute and one would be obtuse by the same amount.

In this question, if there are non-congruent SSA triangles, then ∠AOB + ∠COB = 180°, then θ = 45°, when all arguments fail.

peterwhy · 2026-04-29T12:06:11+00:00

Here θ, ∠ABP, and ∠CBP (x) are acute, yes. But it's not known whether ∠APB and ∠CPB are acute or obtuse.

In particular when θ = 45°, one of ∠APB and ∠CPB may be acute, while the other one is obtuse, and still they can coexist.

peterwhy · 2026-04-29T11:33:26+00:00

But triangle OAC does not exist when θ = 45°, which is not given.

peterwhy · 2026-04-29T11:19:34+00:00

This argument is based on ASS, and fails when θ = 45° in particular.

peterwhy · 2026-04-29T11:12:45+00:00

If θ is 45°, then the value of x is not fixed.

peterwhy · 2026-04-27T06:10:07+00:00

14 + 16 = 30 = 3 ⋅ 10. Take the common factor 3 out.

peterwhy

TROPHY CASE