Snowy panda. Note super cool neck gaiter from back in the states.

Wow, so it’s snowing a lot tonight. This picture doesn’t do it justice – this is the snow that was left on my body after I put away my bike, walked up three flights of stairs, fished around for my keys, got in my apartment, took off my shoes, put away my perishable groceries and then hunted around for my camera. Believe me, I looked like the abominable snowpanda before that!!

Nothing else of interest to add tonight, but tomorrow I’ve made up my mind to do an expose on the mysterious rolls of toilet paper that seem to just be multiplying like rabbits in my classrooms. Does anyone else know what I’m talking about…?

In honor of the snowy weather, check out this cool site! You can make your own snowflake!! See, we should have done things like this back in middle school geometry. Then I’m certain I would have liked math a lot more in high school and college. But nooooo…!! We had to do proofs.

A

//\

/ / /|\

/ / / | \

/ / / | \

/ / / | \

C— — —/—-/—-|—-\B

M K D

Given: triangle ABC; altitude AD; angle bisector AK; median AM angle CAM=MAK=KAD=DAB (all the four smaller angles forming BAC are equal)

Prove: angle BAC is a right angle

Geometry Proof< FONT> Panda Proof

Triangles ADK and ADB are congruent by ASA, so KD=DB.

Also, CM=MB by the definition of a median, so CD = 2 MD + KD.

Let CAM=MAK=KAD=DAB=x.

Then 2 tan 2x + tan x = 2 (MD/AD) + (KD/AD) = CD/AD = tan 3x.

Since tan 2x = (tan 3x – tan x)/(1 + tan x tan 3x), substition yields either tan x = tan 3x or tan x tan 3x = 1, i.e., either tan x = tan 3x or tan 3x = tan (pi/2 – x).

Since x is between 0 and pi/4 exclusive, the first alternative is impossible and the second implies that 4x = pi/2, so BAC is right, as desired. Triangles ADK and ADB sorta’ look the same.

Therefore, if we use our pencils as make shift rulers, we can kind of measure to make sure than CAM is roughly equivelant to KAD and DAB.

*furious scribbling*

Then umm…. uhh… ummmm…

*looks in the back of the book*

Ah-ha!! See, there you go! multiply by some thingamajing, smudge your pencil a bit so the teacher can’t read it, and then very obviously BAC is what you’re after!

Wow, I hated math.

Now listening to: “BT – Never gonna come back down”

(Green is like a boom to the what’s dis non. / Diddy on dawn to the don don diggy dawn…

Yes, Wantin, you’re right, you did send me this song…)

10:30 am

Wow, so it’s snowing a lot tonight. This picture doesn’t do it justice – this is the snow that was left on my body after I put away my bike, walked up three flights of stairs, fished around for my keys, got in my apartment, took off my shoes, put away my perishable groceries and then hunted around for my camera. Believe me, I looked like the abominable snowpanda before that!!

Nothing else of interest to add tonight, but tomorrow I’ve made up my mind to do an expose on the mysterious rolls of toilet paper that seem to just be multiplying like rabbits in my classrooms. Does anyone else know what I’m talking about…?

In honor of the snowy weather, check out this cool site! You can make your own snowflake!! See, we should have done things like this back in middle school geometry. Then I’m certain I would have liked math a lot more in high school and college. But nooooo…!! We had to do proofs.

A

//\

/ / /|\

/ / / | \

/ / / | \

/ / / | \

C— — —/—-/—-|—-\B

M K D

Given: triangle ABC; altitude AD; angle bisector AK; median AM angle CAM=MAK=KAD=DAB (all the four smaller angles forming BAC are equal)

Prove: angle BAC is a right angle

Geometry Proof< FONT> Panda Proof

Triangles ADK and ADB are congruent by ASA, so KD=DB.

Also, CM=MB by the definition of a median, so CD = 2 MD + KD.

Let CAM=MAK=KAD=DAB=x.

Then 2 tan 2x + tan x = 2 (MD/AD) + (KD/AD) = CD/AD = tan 3x.

Since tan 2x = (tan 3x – tan x)/(1 + tan x tan 3x), substition yields either tan x = tan 3x or tan x tan 3x = 1, i.e., either tan x = tan 3x or tan 3x = tan (pi/2 – x).

Since x is between 0 and pi/4 exclusive, the first alternative is impossible and the second implies that 4x = pi/2, so BAC is right, as desired. Triangles ADK and ADB sorta’ look the same.

Therefore, if we use our pencils as make shift rulers, we can kind of measure to make sure than CAM is roughly equivelant to KAD and DAB.

*furious scribbling*

Then umm…. uhh… ummmm…

*looks in the back of the book*

Ah-ha!! See, there you go! multiply by some thingamajing, smudge your pencil a bit so the teacher can’t read it, and then very obviously BAC is what you’re after!

Wow, I hated math.

Now listening to: “BT – Never gonna come back down”

(Green is like a boom to the what’s dis non. / Diddy on dawn to the don don diggy dawn…

Yes, Wantin, you’re right, you did send me this song…)

10:30 am

A little off-topic but just wanted to say I liked the layout of the site